Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
|---|---|---|
Counting how many ways 30 green balls... How many ways are there to distribute 30 green balls to 4 persons if Alice and Eve together get no more than 20 and Lucky gets at least 7?
The answer given to me was $2464 = C(26, 3) − 66 − 46-24$ but I got $C(26, 3) - 6$.
Here is what I did:
Using "stars and bars" (or whatever the actual name for it is), I found the total number of possible combinations. Since 7 balls must immediately go to Lucky, there are 23 balls left to distribute. There are 3 'bars', so the total number of possible combinations is $\binom{26}{3}$.
Because this number accounts for the situations where Alice and Eve have $
\geq 21$ balls, I must subtract those out. (Here is where I think I messed something up).
When Alice and Eve have 21: This leaves 2 balls left to distribute, so $\binom{3}{2}$
When Alice and Eve have 22: This leaves 1 ball left to distribute, so $\binom{2}{1}$
When Alice and Eve have 23: This leaves no balls left to distribute, so it equals 1.
So $C(26, 3) - 6$.
Where did I go wrong?
| Lucky gets $7$ balls right away. Alice and Eve can split their $j\in[0..20]$ balls in $j+1$ ways, then Lucky and Mike can split the remaining $23-j$ balls in $24-j$ ways. The total number of admissible allocations therefore is
$$\sum_{j=0}^{20}(j+1)(24-j)=2464\ .$$
| {
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An open set which does not belong to Uniform Topology of $R^w$ but belongs to $l^2$ topology of $R^w $ I can not find an open set which does not belong to Uniform Topology of $R^w$ but belongs to $l^2$ topology of $R^w $.
I know that $l^2$ topology of $R^w $ contains Uniform Topology of $R^w$. But I can not find those sets which make $l^2$ topology of $R^w $ bigger than Uniform Topology of $R^w$.
Can anyone please help me?
| As Nate Eldredge(https://math.stackexchange.com/users/822/nate-eldredge) mentioned , the unit ball i.e the set of all sequences $x$ satisfying $∑|x(n) |^2<1$ can be an example.
There are sequences of arbitrarily small uniform norm which are not contained in this ball.
So this is an open ball around the zero sequence in $l^2$ which is not so in the Uniform Topology. As open ball around zero sequence of radius $1$ will look like $B(0,1)=⋃_{δ<1 }U(0,δ)$ where $U(0,δ) = (-δ , δ) \times (-δ , δ) \times...........$
.
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The Modulus of all the roots of a Polynomial are equal to $1$
Suppose the real number $\lambda \in (0,1)$, and let $n$ be a positive integer. Prove that all roots of the polynomial $$f\left ( x \right )=\sum_{k=0}^{n}\binom{n}{k}\lambda^{k\left ( n-k \right )}x^{k}$$ have modulus equal to $1.$
The Putnam problem 2014 B4 is similar: Show that for each positive integer $n,$ all the roots of the polynomial $\sum_{k=0}^n 2^{k(n-k)}x^k$ are real numbers.
| Write the above polynomial in the form: $a+bx+cx^2$... and observe that if we put $f(x)=0$ it is same as putting $x^k f(1/x)=0$ {property of binomial coefficients and the power of lambda} conclude that if $x$ is a root then $1/x$ is also has the same modulus. Hence $|x|=1$
| {
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Image of a basis forms a basis, if and only if matrix is invertible Suppose $B_1=\{v_1,v_2,...,v_n\}$ is a basis of $\mathbb{R}^n$, and $M$ is an $n*n$ matrix. Prove that $B_2=\{Mv_1,Mv_2,...,Mv_n\}$ is also a basis of $\mathbb{R}^n$ if and only if $M$ is invertible.
Following is what I have so far:
Assume $B_2$ is basis of $\mathbb{R}^n$.
Then, $B_2$ is a set of linearly independent vectors, and $B_2$ spans $\mathbb{R}^n$.
Since $B_1$ is also a basis of $\mathbb{R}^n$, then any element(vector) of $B_2$ is a linear combination of elements(vectors) of $B_1$ and vice-versa.
$Mv_1= a_{11}v_1+a_{21}v_2+...+a_{n1}v_n$ , where $a_{11},a_{21},...,a_{n1}\in \mathbb{R}$
Likewise, $Mv_2= a_{12}v_1+a_{22}v_2+...+a_{n2}v_n$ , where $a_{12},a_{22},...,a_{n2}\in \mathbb{R}$
$\begin{bmatrix}Mv_1&Mv_2&...&Mv_n\end{bmatrix}=\begin{bmatrix}v_1&v_2&...&v_n\end{bmatrix}\begin{bmatrix}a_{11}&a_{12}&...&a_{1n}\\a_{21}&a_{22}&...&a_{2n}\\ \vdots&\vdots&\vdots&\vdots\\a_{n1}&a_{n2}&...&a_{nn}\end{bmatrix}$
Not sure what to do next ...
|
$\begin{bmatrix}Mv_1&Mv_2&...&Mv_n\end{bmatrix}=\begin{bmatrix}v_1&v_2&...&v_n\end{bmatrix}\begin{bmatrix}a_{11}&a_{12}&...&a_{1n}\\a_{21}&a_{22}&...&a_{2n}\\ \vdots&\vdots&\vdots&\vdots\\a_{n1}&a_{n2}&...&a_{nn}\end{bmatrix}$
The matrix on the right is just $M^T$.
Suppose you have a linear combination of the original basis u = [$v_1,v_2...v_n$]$[c_1, c_2 ... c_n]^T$
And suppose you're trying to find a linear combination of the new basis
u = [$Mv_1,Mv_2...Mv_n$]$[c'_1, c'_2 ... c'_n]^T$
We can rewrite this as
u = [$v_1,v_2...v_n$]$M^T[c'_1, c'_2 ... c'_n]^T$ =
[$v_1,v_2...v_n$]$([c'_1, c'_2 ... c'_n]M)^T$
So we can simply set $[c'_1, c'_2 ... c'_n] = [c_1, c_2 ... c_n]M^{-1}$
| {
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Matrix square roots of -I Since we can see matrices as generalizations of complex numbers, I asked myself if there is a way to classify those matrices which are the "Basis" for the complex part. That is, I would like to identify the set of $n\times n$ real valued matrices $M$ whose square $M^2$ is equal to $-I$, where $I$ is the $n\times n$-identity matrix. Is this set already classified?
The matrix $J = \left( \begin{smallmatrix}0 & -1\\1 & 0 \end{smallmatrix} \right)$ satisfies $J^{2} = -I$ in the 2-dimensional case.
EDIT:
Thanks to your comments and answers I have the following observation (if its wrong, please tell me):
Suppose that $M$ satisfies that $M^t=-M$, then we can observe for two vectors $x,y\in\mathbb{R}^{2n}$ that
$y^tAx=(Ax)^{t}y=x^{t}A^ty=-x^tAy$
My interpretation is that if the angle between $x$ and $Ay$ is $\alpha$ then $\pi+\alpha$ is the angle between $Ax$ and $y$, since ${\displaystyle \cos \;x=-\cos(x+\pi )}$ (here we suppose that the length of the vectors are not relevant). If now $y=x$ and then we can see that $Ax$ is orthogonal to $x$.
Is it true that $A^2$ will be a rotation of 180 degrees, right? Because of vector length we will have something like $A^2=-CI$, where $C$ is just a constant. Is that true?
| You might be interested in complex structures on real vector spaces.
The basic idea is this: suppose I have a complex vector space $V$ of dimension $n$. If I forget how to multiply by $i$, and only use scalars that are real, then $V$ is also a real vector space of dimension $2n$ (let's call this $V_{\mathbb{R}}$, to make it clear I've forgotten anything complex). An interesting question is: what exactly did I forget when I forgot how to multiply by $i$? Can I remember it for later, and get back my original complex vector space?
On the complex vector space $V$, there is a linear map $J: V \to V$ which is multiplication by $i$, defined by $Jv = iv$. In fact, this map is $\mathbb{R}$-linear, and so gives a linear map $J: V_\mathbb{R} \to V_\mathbb{R}$. We can check that $J^2v = i^2v = -v$, and so $J^2 = -\mathrm{id}_V$. This map $J$ is exactly what we forgot when we forgot how to multiply by $i$: if I have a vector $v \in V_{\mathbb{R}}$ and want to scale it by $(a + bi)$, then I use $(a \,\mathrm{id}_V + bJ)v$.
Your question is essentially going the opposite way: starting with a $2n$-dimensional real vector space $W$, what are the kind of maps $J: W \to W$ such that $(W, J)$ becomes a complex vector space with action $(a + bi)w = (a \, \mathrm{id}_W + bJ)w$.
| {
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Period of the pendulum and taylor expansion The period of a (non-linear) simple pendulum is
$$
T(\theta_0) = \sqrt{8}/\omega_0 \int_0^{\theta_0} \frac{1}{\sqrt{\cos\theta-\cos\theta_0}}d\theta.
$$
Using elliptic functions, we can show that the term of order $1$ in $\theta_0$ is $2\pi/\omega_0$, which is precisely the period of the linear simple pendulum.
On the other hand, writing
$$
\cos\theta-\cos\theta_0 \simeq \sin\theta_0(\theta_0-\theta)
$$
leads to
$$
T(\theta_0) \simeq \sqrt{5}/2/\omega_0\simeq 5.65/\omega_0
$$
which is far from the result $2\pi/\omega_0$.
Is it due to the fact I neglected terms of order $>1$?
So if I would take the Taylor-Lagrange expansion
$$
\cos\theta-\cos\theta_0=(\theta_0-\theta)\sin\theta_0-1/2(\theta_0-\theta)^2\cos \xi, \quad \xi\in (0,\theta_0),
$$
could I compute $\xi(\theta_0)$ so that
the first term of $T(\theta_0)$ in $\theta_0$ is $2\pi/\omega_0$ ?
| The issue is that $$\frac{\cos\theta-\cos\theta_0}{\theta-\theta_0}\approx -\sin\theta_0$$ only if $\theta$ and $\theta_0$ are close, otherwise that is not a good approximation, so you cannot recover the exact first term of the wanted Taylor series from it, since the value of the actual integrand function in a right neighbourhood of the origin is fairly different from the value of the approximated integrand function.
On the other hand
$$\frac{\cos\theta-\cos\theta_0}{\theta^2-\theta_0^2}\approx -\frac{\sin\theta_0}{2\theta_0}$$
leads to the correct outcome since $\int_{0}^{T}\frac{dt}{\sqrt{T^2-t^2}}=\frac{\pi}{2}$ for any $T>0$.
| {
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On the asymptotic behavior of the Fourier coefficients of $(1-R\cos\theta)^{-3/2}$ Today in class I showed some ways for dealing with the classical integral $\int_{0}^{2\pi}\frac{d\theta}{(A+B\cos\theta)^2}$ under the constraints $A>B>0$, including
*
*Symmetry and the tangent half-angle substitution;
*Relating the integral to the area enclosed by an ellipse, via the polar form of an ellipse with respect to a focus and the formula $\text{Area}=\pi a b =\frac{1}{2}\int_{0}^{2\pi}\rho(\theta)^2\,d\theta$;
*Computing the geometric-like series $\sum_{n\geq 0} r^n \sin(n\theta)$ and applying Parseval's identity to it;
*Applying Feynman's trick (differentiation under the integral sign) to $\int_{0}^{2\pi}\frac{d\theta}{1-R\cos\theta}$ which is an elementary integral due to point 1.
I finished the lesson by remarking that the point $4.$ allows to compute $\int_{0}^{2\pi}\frac{d\theta}{\left(1-R\cos\theta\right)^3}$ almost without extra efforts, while the $L^2$ machinery (point 3.) does not seem to grant the same. Apparently I dug my own grave, since someone readily asked (with the assumption $R\in(-1,1)$)
What is the asymptotic behavior of the coefficients $c_n$ in
$$ \frac{1}{\left(1-R\cos\theta\right)^{3/2}}= c_0+\sum_{n\geq 1}c_n \cos(n\theta) $$
? (Q2) Do we get something interesting by following the "unnatural" approach of applying Parseval's identity to such Fourier (cosine) series?
At the moment I replied that the Paley-Wiener theorem ensures an exponential decay of the $c_n$s, and with just a maybe to the second question. Later I figured out a pretty technical proof (through hypergeometric functions) of
$$ |c_n| \sim K_R\cdot\sqrt{n}\cdot\left(\frac{|R|}{1+\sqrt{1-R^2}}\right)^n \quad \text{as }n\to +\infty$$
and the fact that $c_n$ is given by a linear combination of complete elliptic integrals of the first and second kind. (Q1) I would like to know if there is a more elementary way for deriving the previous asymptotic behaviour. And the outcome of (Q2).
| I don't think this is more elementary, but it may be a little more straightforward. Start with Heine's toroidal identity,
$$ \frac{1}{\sqrt{z-\cos\,t}} = \frac{\sqrt{2}}{\pi} \sum_{m=-\infty}^\infty\,Q_{m-1/2}(z)\,\exp{(i \, m\, t}), \, \, |z|\ge1, \, 0\le t \le 2\pi \, .$$
Splitting the summation into positive and negative $m$ and using the fact that for the toroidal function (Gradshteyn & Rhyzhik 8.737.4)
$Q_{-(m-1/2)-1}(z) = Q_{m-1/2}(z)$ for integer $m$ then one obtains
$$ \frac{1}{\sqrt{z-\cos\,t}} = \frac{\sqrt{2}}{\pi}\Big(2 \sum_{m=1}^\infty\,Q_{m-1/2}(z)\,\cos{( m\, t)}+ Q_{-1/2}(z)\Big) .$$
Differentiate both sides with respect to $z$ and use G&R 8.706.1 (with the fact that $z>1$ to get the phase correct) to get
$$ (z-\cos\,t)^{-3/2} = \frac{-2\sqrt{2}}{\pi\,\sqrt{z^2-1}}\Big(2 \sum_{m=1}^\infty\,Q_{m-1/2}^{\,1}(z)\,\cos{( m\, t)}+ Q_{-1/2}^{\,1}(z)\Big) .$$
From the problem's statement, with $z=1/R,$ we see that the associated toroidal function $Q_{m-1/2}^{\,1}(z)$ must be asymptotically estimated. G&R 8.777.2 has a hypergeometric representation that has an asymptotic nature for large $m,$
$$Q_{m-1/2}^{\,1}(z)=-2\sqrt{\pi}\,\frac{\Gamma(m+3/2)}{\Gamma(m+1)}
\,\frac{\sqrt{z^2-1}}{\zeta^{\,m+3/2}} F\big(3/2,m+3/2; m+1; 1/\zeta^2\big)$$
where $\zeta = z+\sqrt{z^2-1} > 1.$ For convenience set $y=1/\zeta^2$ and expand the ration of Pochhamer symbols in the hypergeometric function as
$$\frac{(m+3/2)_k}{(m+1)_k} \sim 1 + \frac{k}{2m} + \, ... \,\, m\to \infty.$$ Therefore
$$F\big(3/2,m+3/2; m+1; 1/\zeta^2\big)\sim \sum_{k=0}^\infty \dfrac{(3/2)_k}{k!} \big(1+\frac{k}{2m}\big)y^k= (1-y)^{-3/2}\Big(1+\frac{3y}{4m(1-y)} \Big).
$$
For this to be a useful asymptotic expression, $y$ needs to bounded away from 1, which implies $z$ must be bounded from 1, and $R$ must be bounded from 1.
Writing
$$ (z-\cos\,t)^{-3/2} =d_0(z)+\sum_{m=1}^\infty d_m(z)\,\cos{(m\,t)} $$
then, using only the first term of the expression above and the first in the asymptotic expansion of the ratio of gamma functions,
$$ d_m(z) \sim \frac{8\sqrt{2\, m}}{\sqrt{\pi}} (z+\sqrt{z^2-1})^{-3/2}(1-y)^{3/2}$$
In the proposer's notation it is clear to see that
$$c_m = R^{-3/2}d_m(z \to 1/R)$$ and, with some simplification one finds
$$c_m \sim \frac{4}{\sqrt{\pi}} \,\sqrt{m}\, (1-R^2)^{-3/4}
\Big(\frac{R}{1+\sqrt{1-R^2}}\Big)^m .$$
This expression was derived for $z>1$ implying $0\le R<1.$ To extend to
$-1\le R \le 0,$ simply let $t \to t+\pi$ and a factor of $(-1)^m$ is introduced into the summand. It is then clear that the $(-1)^m$ can be incorporated into the final expression as the proposer did, by making the unsigned $R$ an absolute value.
I have checked many of the intermediate steps in Mathematica, but to do so one must use $Q_{m-1/2}^{\,1}(z)=$ LegendreQ[m-1/2, 1, 3, z].
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Abelian group quotient computation New to group theory. Is the following correct? $a, b, c, d$ are independent elements generating the free abelian group $A = \langle a, b, c, d\rangle = \mathbb{Z}a \oplus \mathbb{Z}b \oplus \mathbb{Z}c \oplus \mathbb{Z}d$, and $B$ is the subgroup $B = \langle a, 2b-a, 2c-b, 2d-c\rangle$.
$$
A / B \cong \mathbb{Z}/2 \oplus \mathbb{Z}/4 \oplus \mathbb{Z}/8
$$
Thank you!
| In $A/B$, we have $c=2d$, $b=2c=4d$, $a=2b=8d$, $a=0$.
Therefore, $A/B$ is generated by the class of $d$ and this class has order $8$. Thus, $A/B \cong \mathbb Z/8 \mathbb Z$.
An explicit isomorphism is induced by the map $A \to \mathbb Z/8 \mathbb Z$ given by $\alpha a + \beta b + \gamma c + \delta d \mapsto 4 \beta + 2\gamma + \delta \bmod 8$.
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A $5\times 5$ grid of single-digit numbers in $\mathbb N$, with one cell empty. What number should be in the cell?
16.$$\begin{array}{|c|c|c|c|c|}
\hline
2 & 7 & 4 & 3 & 5
\\\hline
7 & 3 & 4 & 5 & 4
\\\hline
1 & 3 & 2 & 2 & 6
\\\hline
2 & 4 & 5 & 4 & \mathbf{?}
\\\hline
8 & 3 & 6 & 3 & 5
\\\hline
\end{array} $$
Is the answer 1, 2, 3, 4, or 5?
Photograph of the problem source
Really I don't understand. How does such a question relate to logic? For me, it's a game about numbers.Not Logic. It really annoys me to solve such a question. Anyway, I took 12 minutes for this question in exam. I came home. I could not even "solve" it at home ether. I think, such a question is nonsense. No science has anything to do with it. Please help me with the question and please explain me, what does it really mean to solve such a question?
|
$$\begin{array}{c}
x \\
y \\
p \\
q \\
r
\end{array} \qquad\to\qquad x^y = pqr\quad\text{(concatenated)} $$
"[W]hat does it really mean to solve such a question?"
Well, such a question challenges us to find order amid chaos. That's what mathematics ---as the study of pattern--- is all about, so it's not a completely irrelevant mental exercise. As others mention, it could be instructive to find ways to justify any answer. (It's always possible to do that in a puzzle like this one, although not all rules are particularly "nice" ... but "nice" is subjective and not actually required.)
That said, this kind of thing is a horrible exercise for an exam. Not only is it unreasonable to expect someone to notice any pattern in a fixed time-frame, it is unreasonably-unreasonable to expect someone to notice the (so-called) "correct" pattern at all, since that's often akin to mind-reading. ("What was the author thinking?")
In any case, here's a walk-through of my thought process: After a minute or so of skimming the grid, and just before abandoning the whole thing, I happened to recognize the powers "27", "128", and "256" amid the columns of digits; then ---oh, yeah!--- "243" and "343" (which aren't always on the tip of my brain); and then ---hey!--- that could be "625" in the last column! I must be onto something! Yet ... "73" and "44" and friends aren't powers, so maybe "27" was a red herring. Then, the "aha!" moment: $2^7 = 128$. Done.
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Consider the ideal $I = \langle(3,4) \rangle$ of the ring $\mathbb{Z} ×\mathbb{Z}$. Prove that $(\mathbb{Z}×\mathbb{Z})/I$ is not a domain. Having some trouble with this. Need to show that there exist some Zero Divisor, but not really sure what that would look like in $(\mathbb{Z}×\mathbb{Z})/I$
Thanks in advance
| Hint: $(3,1) \cdot (1,4) = (3,4) \in I$.
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Direct sum of two subspace Given the definition on textbook: Let $V$ be a subspace of $\mathbb R^n.$ Every vector $u \in \mathbb R^n$ can be written uniquely as $u = n + p.$ I still don't understand what it means because i am ask to do with a question that ask me the prove the uniqueness claim in the definition.
The question is asking:
Let $v\in \mathbb R^n$. Suppose $p_1, p_2 \in V$ and $n_1,n_2 \in V^\bot$ are such that $v=p_1+n_1 = p_2 + n_2.$ Prove that $p_1 = p_2$ and $n_1 = n_2$.
Anyone can help me answer to this question?
| Let's look at an easy example.
Let $n=2$ and $V= <\begin{pmatrix} 1\\0 \end{pmatrix}>$. Then $V^{\perp}=<\begin{pmatrix} 0\\1 \end{pmatrix}>$. Now pick some $x\in\mathbb{R}^2$, say, $x=\begin{pmatrix} x_1\\ x_2 \end{pmatrix}$. Then $x=n+p$ where $n=x_1\begin{pmatrix} 1\\0 \end{pmatrix}\in V$ and $p=x_2\begin{pmatrix} 0\\1 \end{pmatrix}\in V^{\perp}$. The theorem tells you that these are your only choices of $n$ and $p$ such that $x=n+p$.
Now to the proof:
Let $v=n_1+p_1$ and $v=n_2+p_2$. where $n_i\in V$ and $p_i\in V^{\perp}$ for $i=1,2$. Then $$0=(p_1-p_2)+(n_1-n_2)$$. Since $V$ and $V^{\perp}$ are subspaces of $\mathbb{R}^n$, we have that $p_1-p_2\in V$ and $n_1-n_2\in V^{\perp}$. By definition of $V$ and $V^{\perp}$, we get that $p_1-p_2 \perp n_1-n_2$. But if two vectors are orthogonal, they must be linearly independent. Hence we have that 0 is written as a linear combination of two linearly independent vectors, hence these two vectors must be zero, yielding $p_1-p_2=0$ and $n_1-n_2=0$ which proves that they were the same.
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Prove using mathematical induction $1\cdot2+2\cdot2^2+3\cdot2^3+\ldots+n\cdot2^n=2[1+(n-1)2^n]$
Prove the result using Mathematical Induction $$1\cdot2+2\cdot2^2+3\cdot2^3+\ldots\ldots+n\cdot2^n=2[1+(n-1)2^n].$$
I've been stuck on this problem for hours, I have no idea how do even calculate it. The exponents throw me off. If anyone can help me break it down step-by-step, I would truly appreciate it.
Here's my attempt
| Hint:
The base case is fairly easy: $0\times\left(\ldots\right)=2\left(1+(0-1)\times2^0\right)$. If we assume that $\sum_{n=0}^k n2^{n}=2\left(1+(k-1)\cdot2^{k}\right)$, then our goal for the induction case is to prove that: $$
\underbrace{2\left(1+(k-1)\cdot2^{k}\right)}_{\sum_{n=0}^k {n2^{n}}}+ \underbrace{(k+1)2^{k+1}}_{\text{$(k+1)$'th term}}=2\left(1+k\cdot2^{k+1}\right)$$
Then, divide through by 2 and expand. Can you take it from here?
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If $A$ is idempotent then $A$ is similar to a diagonal matrix with only $0$'s and $1$'s on the diagonal. I am trying to use Jordan normal form to show that if $A^2 = A$ then it is similar to a diagonal matrix with only $0$'s and $1$'s. I've proved that the eigenvalues of $A$ have to be either $0$ or $1$ so we know the diagonal elements of the JNF have to be $0$ or $1$. How do we know that none of the off-diagonals above are $1$?
I don't full understand JNF - I've only learned about it in terms of elementary divisors and the minimal and characteristic polynomials, but a lot of resources online talk about it in terms of eigenspaces which is confusing.
| If $A^2=A$, then the equality should hold for each of the Jordan blocks. In more detail, $A=SJS^{-1}$. Then
$$
SJS^{-1}=A=A^2=SJ^2S^{-1},
$$
and then $J^2=J$. Now you can compare the individual Jordan blocks. In a Jordan block $J_1$, that is not diagonal and with eigenvalue $\alpha$, the $1,2$ entry is $1$. In $J_1^2$, the $1,2$ entry is $2\alpha$. So you would need $2\alpha=1$, which cannot happen here because $\alpha$ is either $0$ or $1$.
| {
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I need the steps to solving this limit without using l´Hopital rule I've tried many ways of solving this limit without using l'Hopital and I just can't figure it out. I know the answer is $3/2 \sin (2a).$
$$\lim_{x\,\to\,0} \frac{\sin(a+x)\sin(a+2x) - \sin^2(a)} x$$
Thank you!
| Perhaps not the most elegant, but:
$\lim_\limits{x\to0} \frac{\sin(a+x)\sin(a+2x) - \sin^2(a)} x\\
\lim_\limits{x\to 0} \frac{\frac 12 (-\cos(2a + 3x) + cos x) - \sin^2(a)} x\\
\lim_\limits{x\to 0} \frac{\frac 12 (-\cos(2a)\cos 3x + \sin(2a)\sin (3x) + \cos x) - \sin^2(a)} x\\
\lim_\limits{x\to 0} \frac{\frac 12 (-\cos(2a)\cos 3x + \sin(2a)\sin (3x) + \cos x) - \frac 12 (1-\cos 2a)} x\\
\lim_\limits{x\to 0} \frac{-\cos(2a)\cos 3x + \sin(2a)\sin (3x) + \cos x - 1 +\cos 2a} {2x}\\
\lim_\limits{x\to 0} \frac{\cos(2a)(1-\cos 3x)}{2x} - \frac {1-\cos x}{2x} + \frac { \sin(2a)\sin (3x)}{2x}$
The first term evaluates to $0,$ the second term evaluates to $0,$ the third term evaluates to $\frac 32 \sin 2a$
| {
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Simple Exclusion Process on a $100 \times 100$ checkerboard This is a problem given by a professor that has been perplexing me.
Suppose a particle takes a random walk on a $100 \times 100$ checkerboard in the following way. After an exponential time with rate 1, it attempts to move up, down, left, or right -- each with probability 1/4. If the attempted move would take the particle off the board, it stays put instead. Then, after an exponential time with rate 1, it tries to move again, and on and on. What is the stationary distribution of the particle's position.
Now suppose there are 1278 such particles on the board moving independently, and multiple particles can occupy the same squares. What is the stationary distribution for the number of particles on each square? You might want to think of your state space as consisting of all the $100 \times 100$ arrays, where the number in the $(i, j)$ position in the array corresponds to the number of particles there.
Finally answer the previous question when the 1278 particles are only allowed to move to empty squares. That is, each square can only accomodate one particle. Now the state space would be all the $100 \times 100$ arrays of 1s and 0s with exactly 1278 1s.
Now, I'm certain that if I understand how to solve the first part (the first paragraph that is, then I'll be able to use similar logic to solve the latter paragraphs. I believe that the principle I need to use because this process is clearly time reversible is
$$P_i^A = \frac{P_i}{\sum_{j \in A}P_j}$$
where $A$ is a truncated state space of the whole CTMC.
I can envision truncating the space to just the neighboring checkerboard squares, but I'm still uncertain of what to even do with that information.
| A clever approach for the first part is to consider an infinite board ruled off in $100 \times 100$ sections. When the particle moves off the edge of your board, it moves to a space of the same type (corner or edge) of a neighboring board, but on the infinite grid all the cells are equivalent as there are no reflections. This shows the chance of the particle being on any given cell on your board is the same. For the second part, you have $1278$ particles each with a chance of $10^{-4}$ to be in your given cell.
| {
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Strange maths coincidence $(6\times 9)+(6+9)=69?$ Is this a freak coincidence in maths or there are more of this type of maths calculation?
$$\color{red}{(6\times 9)}+\color{blue}{(6+9)}=69$$
I try to find more, but I can't. Can you?
| $$(6 \times 9) + (6 + 9) = 10 \times 6 + 9 = 69$$
Similarly,
$$(n \times 9) + (n + 9) = 10 \times n + 9 = \text{"}n9\text{''},$$
for any one-digit number $n$
| {
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If $\forall x,y \in B$ if $(x,y) \in S$ and if $x+y$ is even then $x=y$ then each class of $S$ has at most $2$ elements
Set $B = \{1,2,3,4,5\}$, $S$ - equivalence relation. It is given that for all $x,y \in B$ if $(x,y)\in S$ and if $x+y$ is an even number then $x = y$. In such case is it true that:
*
*the number of elements in each equivalence class of $S$ is at most $2$
*any relation $S$ would have an equivalence class made up of just one even number.
As far as I understand $S$ could be only of such form:
$$
S = \begin{pmatrix}1&2&3&4&5\\1&2&3&4&5 \end{pmatrix}
$$
because all pairs of $(x,y), x \neq y$ which are both odd numbers can't be in $S$ as well as all pairs $(x,y), x \neq y$ which are both even numbers for example:
$$
\begin{pmatrix}1\\3 \end{pmatrix}, \begin{pmatrix}2\\4 \end{pmatrix}
$$
because their sum will be even but $x \neq y$.
In addition $(x,y), x\neq y$ where one of them is odd and one is even also can't be in $S$ because then the relation will not be transitive and hence will not be an equivalence relation.
In such case I think the statement 1 is false because all equivalence classes are exactly of size $1$ and statement 2 is true because we have for example the equivalence class $\{2\}$ which is one even number.
I'm not sure about my logic because the question is quite tricky.
| Assume that $S$ has an equivalence class with $2$ even numbers, namely $2$ and $4$. Then, $(x,y)\in S$ and $x+y=6$ which is even, but this contradicts $x=y$. Hence, two even numbers cannot be in the same equivalence class.
Now, assume that there exists an equivalence class with two odd numbers $x\neq y$. Then $x+y=\text{even},$ but again $x=y$ is not satisfied. Hence, if $x\neq y$ and $x,y$ odd, then they cannot be in the same equivalence class.
So, you have that any equivalence class in $S$ contains at most one even and at most one odd number. Since, there are five elements in $B$, this leaves you with a class that necessarily has one element. But I do not see why this has to be an even number. So, I agree with the first statement but disagree with the second statement (the even numnber part).
| {
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Finding eigenvalues and eigenvectors and then determining their geometric and algebraic multiplcities I have the following matrix:
$A = \begin{bmatrix}
1 && 7 && -2 \\
0 && 3 && -1 \\
0 && 0 && 2 \end{bmatrix}$
and I am trying to find the eigenvalues and eigenvectors followed by their respective geometric multiplicity and algebraic multiplicity.
What I have so far:
$\det(A - \lambda I) = \det\begin{bmatrix}
1-\lambda && 7 && -2 \\
0 && 3-\lambda && -1 \\
0 && 0 && 2-\lambda \end{bmatrix}$
I see that it is an upper triangular matrix so determinant is just the diagonal. Which gives me
$(1-\lambda)(3-\lambda)(2-\lambda)$ which gives me $\lambda = 1,3,2$. I also notice that all three have the algebraic multiplicity of 1 (their exponents were 1).
Following that I move on to the geometric multiplicity:
$ A - 3I = \begin{bmatrix}
-2 && 7 && -2 \\
0 && 0 && -1 \\
0 && 0 && -1 \end{bmatrix}$ which has RRE of $\begin{bmatrix}
1 && -\frac{7}{2} && 0 \\
0 && 0 && 1 \\
0 && 0 && 0 \end{bmatrix}$ which yields
$\begin{bmatrix}
x \\
y \\
z \end{bmatrix} = \begin{bmatrix}
\frac{7}{2}\\
1 \\
0\end{bmatrix} s$ which has a geometric multiplicity of 1.
$ A - 2I = \begin{bmatrix}
-1 && 7 && -2 \\
0 && 1 && -1 \\
0 && 0 && 0 \end{bmatrix}$ which has RRE of $\begin{bmatrix}
1 && 0 && -5 \\
0 && 1 && -1 \\
0 && 0 && 0 \end{bmatrix}$ which yields
$\begin{bmatrix}
x \\
y \\
z \end{bmatrix} = \begin{bmatrix}
5\\
1 \\
1\end{bmatrix} s$ which has a geometric multiplicity of 1.
Finally,
$ A - I = \begin{bmatrix}
0 && 7 && -2 \\
0 && 2 && -1 \\
0 && 0 && 1 \end{bmatrix}$ which has RRE of $\begin{bmatrix}
0 && 1 && 0 \\
0 && 0 && 1 \\
0 && 0 && 0 \end{bmatrix}$
This is where I am stuck, I'm not sure what is the resulting $\begin{bmatrix}
x \\
y \\
z \end{bmatrix}$ that column of 0's is confusing me.
Any help would be appreciated.
Edit: Would I just say that column of zeros is a free variable? Thus giving me $\begin{bmatrix}
1 \\
0 \\
0 \end{bmatrix}$ ? or is that wrong?
| That RRE $\begin{bmatrix} 0&1&0\\ 0&0&1\\ 0&0&0 \end{bmatrix}$ means $y=0$ and $z=0$, so $$\begin{bmatrix} x\\y\\z \end{bmatrix} =\begin{bmatrix} x\\0\\0 \end{bmatrix} =x\begin{bmatrix} 1\\0\\0 \end{bmatrix}.$$
| {
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Method of CDF for Y = 1/X I am trying to solve this question:
Let X be a standard cauchy variable. Define Y to be 1/X. I want to find the CDF of Y.
My problem: I am finding the CDF to be:
https://arachnoid.com/latex/?equ=%5Cfrac%7B1%7D%7B2%20%7D-%5Cfrac%7B1%7D%7B%5Cpi%20%7Darctan(%5Cfrac%7B1%7D%7By%20%7D)
But as a take the limit to infinity, it doesn't equal 1, it equals 1/2. I searched online for answers but I couldn't fix my problem.
| A standard trigonometric identity says
$$
\arctan \frac 1 y = \frac \pi 2 - \arctan y \text{ if } y>0
$$
so we get
$$
\frac 1 2 -\frac 1 \pi \arctan \frac 1 y = \frac 1 2 - \frac 1 \pi\left( \frac \pi 2 - \arctan y \right) \text{ if } y>0
$$
and this simplifies to
$$
\frac 1 \pi \arctan y
$$
Somewhere you should have had $\text{“} + \text{constant''}.$ With the right constant, the limit as $y\to\infty$ is $1.$
And so the density is
$$
\frac 1 {\pi(1+y^2)}.
$$
Then the case where $y<0$ needs to be similarly treated.
Thus you see that if $X$ has a standard Cauchy distribution, then $1/X$ has that same standard Cauchy distribution.
| {
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Legendre polynomials and primality testing Can you provide a proof or a counterexample for the following claim ?
Let $n$ be an odd natural number greater than one . Let $r$ be the smallest odd prime number such that $r \nmid n$ and $n^2 \not\equiv 1 \pmod r$ . Let $P_n(x)$ be Legendre polynomial , then $n$ is a prime number if and only if $P_n(x) \equiv x^n \pmod {x^r-1,n}$ .
You can run this test here .
I have tested this claim up to $2 \cdot 10^4$ and there were no counterexamples .
| See here for more information. This covers the test you are describing, however it is stated as a conjecture, not a primality test.
| {
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Neighbors have three children. Given three independent observations of boy, what is the probability that they have 3 boys? I would really appreciate some help on this one. I'm completely lost. I have no idea why my method doesnt work.
"Your new neighbors have three children. If you are told about three
independent observations of a boy, what is the probability that they
have three boys?"
I solve it by saying that our sample space is $S = \{bbb,ggg,bgb,bbg,ggb,gbg,gbb,bgg\}$.
If we then introduce the event $A_i = \{$observation i of boy$\}$ and call the event $B=\{bbb\}$
then $P(A_i)=7/8 $ for $i=1,2,3$ and the events $A_i$ are independent according to the problem text. Furthermore, clearly $P(B) = 1/8$.
We now want to calculate the probability
$P(B|A_1 \cap A_2 \cap A_3)$ which is (because of total law of probability/Bayes formula) equal to
$\frac{P(B)P(A_1 \cap A_2 \cap A_3 | B)}{P(A_1 \cap A_2 \cap A_3)}$
Now, $P(A_1 \cap A_2 \cap A_3) = P(A_1)P(A_2)P(A_3)$ because of their independence, and I argue that $P(A_1 \cap A_2 \cap A_3 | B) = 1$ because the probability of observing a boy three times given that there only is three boys should be one hundred percent.
This simplifies to
$\frac{8^2}{7^3}$
which is clearly wrong. The answer should be 1/2, but i have no idea how to come up with that.
What am I doing wrong?
Thanks!
| We can restate the question as follows. Suppose there are $3$ balls in an urn, and each ball is either black or white with equal probability. We draw one ball from the urn at random, observe the color, and replace it in the urn. Given that after $3$ draws we observed a black ball each time, what is the probability that the urn contains no white balls?
Let the number of white balls in the urn be the random variable $X$. Then before any data is observed, $X$ is presumed to follow a binomial distribution with parameters $n = 3$ and $p = 1/2$. Let $Y$ represent the number of white balls drawn out of $3$ independent trials with replacement. Then we have $$\Pr[X = 0 \mid Y = 0] = \frac{\Pr[Y = 0 \mid X = 0]\Pr[X = 0]}{\Pr[Y = 0]}$$ by Bayes theorem. Since the prior distribution of $X$ is binomial, we have $$\Pr[X = 0] = \binom{3}{0}(1/2)^0 (1 - 1/2)^{3-0} = \frac{1}{8}.$$ Given that there are no white balls, $\Pr[Y = 0 \mid X = 0] = 1$--we cannot draw a white ball if there is none in the urn. The only remaining quantity to be determined is the unconditional probability of $Y = 0$, which by the law of total probability is simply $$\Pr[Y = 0] = \sum_{x=0}^3 \Pr[Y = 0 \mid X = x]\Pr[X = x].$$ We can already compute the first term of this sum, since it is just the numerator in the desired probability. But the other conditional probabilities when $X \in \{1, 2\}$ require more thought. Specifically, the conditional distribution of $Y \mid X = x$ is binomial with parameters $n = 3$ and $p = x/3$. That $n = 3$ is obvious; $Y$ comprises the sum of the number of white balls obtained from three independent draws with replacement from the urn. The probability of observing a white ball in any single draw is determined by the number of white balls in the urn, which is $x$: so if $x = 0$, all balls are black and $p = 0$, reflecting the impossibility of drawing a white ball. If $x = 1$, then $p = 1/3$ reflects a $1/3$ chance of drawing a white ball. Thus $$\Pr[Y = y \mid X = x]\Pr[X = x] = \binom{3}{y}(x/3)^y (1 - x/3)^{3-y} \cdot \binom{3}{x} (1/2)^x (1 - 1/2)^{3-x} = \frac{1}{216} \binom{3}{x}\binom{3}{y} x^y (3-x)^{3-y}.$$
With this in mind, we compute $$\Pr[Y = 0] = \frac{1}{216} \sum_{x=0}^3 \binom{3}{0} \binom{3}{x} x^0 (3-x)^3 = \frac{54}{216} = \frac{1}{4}.$$ It follows that $$\Pr[X = 0 \mid Y = 0] = \frac{1/8}{1/4} = \frac{1}{2}.$$
From here, the sophisticated student can easily generalize to the case where we are given $n$ balls in the urn, each with a prior probability $p$ of being white, and $m$ iid draws from the urn, in which $y$ are observed white. Then the posterior distribution of the number of white balls is $$\Pr[X = x \mid Y = y] = \frac{\Pr[Y = y \mid X = x]\Pr[X = x]}{\Pr[Y = y]} = \frac{\binom{m}{y} (x/n)^y (1 - x/n)^{m-y} \binom{n}{x} p^x (1-p)^{n-x}}{\sum_{k=0}^n \binom{m}{y} (k/n)^y (1 - k/n)^{m-y} \binom{n}{k} p^k (1-p)^{n-k}}.$$
| {
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Matrix as tensor exercise with answer Can I get an explanation for the following exercise with answer in the book Linear Algebra via Exterior Products by S. Winitzki?
Exercise 1 - Matrices as Tensors. Now suppose you have a matrix $A_{jk}$ that specifies the linear operator $\hat A$ in a basis $\{\mathbf e_k\}.$ Which tensor $A\in V\otimes V^*$ corresponds to this operator?
Answer:
$$A=\sum_{j,k=1}^n\,A_{jk}\,\mathbf e_j\otimes \mathbf e_k^*.$$
A tensor of the form $\mathbf e_j \otimes\mathbf e_k^*$ is presumably a $(1,1)$-tensor, and should take a covector and a vector. But if it represents a matrix, then it is only taking in a vector...
Say the matrix is $$\mathbf A=\begin{bmatrix}6&7&1\\-2&0&10\end{bmatrix}.$$ Then, it would just need a vector to implement the linear transformation. For example, $\small\begin{bmatrix}4\\3\\1\end{bmatrix}.$
Aren't the columns of the matrix $\mathbf A$ expressed in the the basis $\mathbf e_k^*$ as in
$$\mathbf A=\begin{bmatrix}6 \mathbf e_{11}^*&7\mathbf e_{12}^*&1\mathbf e_{13}^*\\-2\mathbf e_{21}^*&0\mathbf e_{22}^*&10\mathbf e_{23}^*\end{bmatrix}$$
in some fashion? What goes with $\mathbf e_j$?
| One of the basic insights of multilinear algebra is that "taking a vector" is in some sense equivalent to "giving you a covector" (because taking a vector is what that covector does), and conversely "taking a covector" is the same as "giving you a vector".
So we can view an $(1,1)$-tensor either as something that takes in a vector and a covector (and gives you a scalar back), or as something that takes a vector and gives you another vector back. Not coincidentally, these are also two things you can do with a matrix: Given $A$ you can either take a column $v$ and a row $z^T$ and compute the scalar $z^T A v$, or you can take just the column $v$ and compute the column $Av$.
In particular, $w\otimes u^*$ can represent the operation that
*
*takes in a vector $v$.
*computes a scalar contracting $u^*$ and $v$.
*returns that scalar times $w$.
If you also give it a covector $z^*$, you can contract that with the result of the above operation and get a scalar. This gives the same scalar as if you contracted the input vector and covector separately:
$$ \langle z^*, \langle u^*,v\rangle w \rangle = \langle u^*, v\rangle\langle z^*, w\rangle $$
by (right) linearity of the inner product.
| {
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Can we find the asymptotic behavior of this $f(x) =\int_{0}^{\infty}\frac{u^2}{1+\frac{e^{u^2}}{x}}du$? I encountered this function in Statistical Mechanics. $$f(x) =\int_{0}^{\infty}\frac{u^2}{1+\frac{e^{u^2}}{x}}du$$
For $x=0$, we define its value to be zero.
I wanted to see it's asymptotic behavior in the limit x tending to $\infty$. Can we express the asymptotic behavior of this function, in the limit x tending to $\infty$, in terms of other known mathematical functions (if possible; in elementary functions)? For some physical reasons (which are irrelevant here), $x$ belongs to $[0,\infty)$.
| 1. (Not so illuminating) analytic expression. Assume for a moment that $0 < x < 1$. Then
\begin{align*}
f(x)
&= \int_{0}^{\infty} \frac{xu^2e^{-u^2}}{1 + xe^{-u^2}} \, du
= \sum_{n=1}^{\infty} (-1)^{n-1} x^n \int_{0}^{\infty} u^2 e^{-nu^2} \, du \\
&= -\frac{\sqrt{\pi}}{4} \sum_{n=1}^{\infty} \frac{(-x)^n}{n^{3/2}}
= -\frac{\sqrt{\pi}}{4} \operatorname{Li}_{3/2}(-x).
\end{align*}
The last function is analytic outside $(-\infty, -1]$, and hence this identity extends to all of $x \geq 0$ by the principle of analytic continuation. But this is not so useful when investigating the asymptotic bahavior of $f(x)$.
2. Asymptotic expansion. Write $\alpha = \log x$ and make the substitution $u = \sqrt{\alpha(v+1)}$. Then
\begin{align*}
f(x)
&= \frac{\alpha^{3/2}}{2} \int_{-1}^{\infty} \frac{\sqrt{1+v}}{1 + e^{\alpha v}} \, dv \\
&= \frac{\alpha^{3/2}}{2} \left( \int_{0}^{1} \sqrt{1-v} \, dv - \int_{0}^{1} \frac{\sqrt{1-v}}{1 + e^{\alpha v}} \, dv + \int_{0}^{\infty} \frac{\sqrt{1+v}}{1 + e^{\alpha v}} \, dv \right).
\end{align*}
This easily yields the following asymptotics
$$ f(x) = \frac{1}{3}(\log x)^{3/2} + \mathcal{O}\left( (\log x)^{1/2} \right). $$
For a better resolution, recall that the polylogarithm is defined as $\operatorname{Li}_s(z) = \sum_{n=1}^{\infty} \frac{z^n}{n^s}$ for $|z| < 1$. Then
$$ \frac{1}{1 + e^{\alpha v}} = -\operatorname{Li}_0(-e^{-\alpha v}), \qquad \frac{d}{dv} \operatorname{Li}_{s+1}(-e^{-\alpha v}) = - \alpha \operatorname{Li}_s(-e^{-\alpha v}) $$
and hence
\begin{align*}
\int_{0}^{\infty} \frac{\sqrt{1+v}}{1 + e^{\alpha v}} \, dv
&= -\int_{0}^{\infty} (1+v)^{1/2} \operatorname{Li}_0(-e^{-\alpha v}) \, dv \\
&= -\frac{\operatorname{Li}_1(-1)}{\alpha} - \frac{1}{2\alpha} \int_{0}^{\infty} \frac{\operatorname{Li}_1(-e^{-\alpha v})}{(1+v)^{1/2}} \, dv \\
&= -\frac{\operatorname{Li}_1(-1)}{\alpha} - \frac{\operatorname{Li}_2(-1)}{2\alpha^2} + \frac{1}{4\alpha^2} \int_{0}^{\infty} \frac{\operatorname{Li}_2(-e^{-\alpha v})}{(1+v)^{3/2}} \, dv
\end{align*}
and, in principle, the same argument can be applied to extract an asymptotic expansion up to any fixed order. Similarly,
\begin{align*}
\int_{0}^{1} \frac{\sqrt{1-v}}{1 + e^{\alpha v}} \, dv
&= -\int_{0}^{1} (1-v)^{1/2} \operatorname{Li}_0(-e^{-\alpha v}) \, dv \\
&= -\frac{\operatorname{Li}_1(-1)}{\alpha} + \frac{1}{2\alpha} \int_{0}^{1} \frac{\operatorname{Li}_1(-e^{-\alpha v})}{(1-v)^{1/2}} \, dv \\
&= -\frac{\operatorname{Li}_1(-1)}{\alpha} + \frac{\operatorname{Li}_2(-1) - \operatorname{Li}_2(e^{-\alpha})}{2\alpha^2} \\
&\qquad + \frac{1}{4\alpha^2} \int_{0}^{1} \frac{\operatorname{Li}_2(-e^{-\alpha v}) - \operatorname{Li}_2(-e^{-\alpha})}{(1-v)^{3/2}} \, dv
\end{align*}
and so on. Using the results above, we obtain a better asymptotics
$$ f(x) = \frac{1}{3} (\log x)^{3/2} + \frac{\pi^2}{24} \frac{1}{(\log x)^{1/2}} + \mathcal{O}\left( \frac{1}{(\log x)^{3/2}} \right). $$
| {
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Consider the string $(∀x)(∀y)(f(x) = y → ((∀z)g(z) = f(x) ≡ (∀z)g(z) = y))$ Is this well-formed formula a tautology?
I am having trouble finding examples to help me work through these types of problems. I was hoping someone could help me out.
I think, using abstraction, this can be written as:
$$(∀x)(∀y)(p → (q ≡ q_1))$$
which can be further abstracted to a simple boolean variable such as $p_1$.
Which I say is not a tautology.
But then I am asked:
Can you prove $$(∀x)(∀y)(f(x) = y → ((∀z)g(z) = f(x) ≡ (∀z)g(z) = y))$$ in predicate logic? If so, give a proof, if not, explain why.
I know that if the above were a tautology, then the above would be immediately provable. But if it is not a tautology, what do I say?
Can I say it is not provable because it is not a tautology and therefore not an axiom... I don't know what to say here.
Thanks in advance for any help or thoughts.
| The formula is a valid formula of First-order logic with equality.
Here is the derivation:
1) $f(x)=y$ --- premise
2) $(∀z)(g(z)=f(x))$ --- assumed [a]
3) $g(z)=f(x)$ --- from 2) by Universal instantiation
4) $g(z)=y$ --- from 1) and 3) bt transitivity of equality
5) $(∀z)(g(z)=y)$ --- from 4) by Generalization
6) $(∀z)(g(z)=f(x)) \to (∀z)(g(z)=y)$ --- from 2) and 5) by Conditional Proof (i.e. Implication introduction)
In the same way, assuming $(∀z)(g(z)=y)$ we derive :
7) $(∀z)(g(z)=y) \to (∀z)(g(z)=f(x))$.
Thus, we derive, by Bi-conditional introduction :
8) $(∀z)(g(z)=y) \leftrightarrow (∀z)(g(z)=f(x))$
and finally we conclude with:
9) $(∀x)(∀y)[(f(x)=y) \to ((∀z)(g(z)=y) \leftrightarrow (∀z)(g(z)=f(x)))]$ --- from 19 and 8) by Conditional introduction (discharging the premise) and Generalization twice.
Having proved it, by soundness of the calculus, we conclude that the formula is valid.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Combining ratios of quantities with different sizes A friend and I had two different answers to this seeming simple question. It goes as follows:
Jar A contains flour and sugar in the ratio 5 : 1. Jar B, which is three times larger than Jar A, contains flour and sugar in the ratio 8 : 1. When the contents of these jars are combined, the resulting mixture contains flour and sugar in the ratio x : 1. What is the value of x?
Thanks for any help. This is not homework of any sorts and we genuinely just want to figure out the proper way of doing it.
| I interpret "three times larger than" to mean "three times plus one the size of" instead of just "three times the size of". So, using a similar reasoning as Duchamp Gérard H. E.:
$$\text{Flour:} \quad (5/6)U+(8/9)4U=(5/6)U+(32/9)U=(79/18)U\\ \text{Sugar:} \quad (1/6)U+(1/9)4U=(1/6)U+(4/9)U=(11/18)U.$$
The ratio $(79/18)$:$(11/18)$ is the same as $x$:$1$, so $x=79/11$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725658",
"timestamp": "2023-03-29T00:00:00",
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Solve $\tan (\theta) + \tan (2\theta) = \tan (3\theta)$
Find the general solution of:
$$\tan (\theta) + \tan (2\theta) = \tan (3\theta)$$
My Attempt:
$$\tan (\theta) + \tan (2\theta) = \tan (3\theta)$$
$$\dfrac {\sin (\theta)}{\cos (\theta)}+ \dfrac {\sin (2\theta)}{\cos (2\theta)}=\dfrac {\sin (3\theta)}{\cos (3\theta)}$$
$$\dfrac {\sin (\theta+2\theta)}{\cos (\theta) \cos (2\theta)}=\dfrac {\sin (3\theta)}{\cos (3\theta)}$$
| I'll give a hint to get you started which is that $\text{tan}(a + b) = \dfrac{\text{tan}(a) + \text{tan}(b)}{1-\text{tan}(a)\text{tan}(b)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove that $\bigcup S_n = (0,1.5)$ if $S_n=\{\frac{1}{n}\le x< 1+\frac{1}{n}\}$?
Given that $S_n=\{x\in \mathbb R|\frac{1}{n}\le x< 1+\frac{1}{n}\}$, $n\in \mathbb N - \{0,1\}$ I need to find $\bigcup S_n$.
If we plug in $n=2$ then $0.5 \le x < 1.5$.
If $n\to \infty$ then $0<x\le 1$. Thefore $\bigcup S_n=(0,1.5)$.
I'm having trouble proving this.
Direction 1: $\bigcup S_n \subseteq (0,1.5)$:
If some $x \in S_n$ then there exists some $m$ such that $\frac{1}{m}\le x<1+\frac{1}{m}$ then because it's a set of unions then $S_n \in \bigcup S_n$.
Direction 2: $(0,1.5)\subseteq \bigcup S_n$:
By Archimedean principle for $x>0$ exists $n\in \mathbb N$ such that $0<\frac{1}{n}<x$.
So if we choose some $n \in \mathbb N-\{0,1\}$ then $0< \frac{1}{n} < 1$ then $(0,1.5)\subseteq \bigcup S_n$.
I'm really not sure about direction 2 and even if I'm right I'd appreciate an explanation of the 2nd direction because I don't really understand it.
| For what it is worth, I was wondering if there was a way to calculate the 1.5 instead of guessing it first, and I found the following proof. This may or may not be to your taste; see e.g. EWD1300 if you are interested in the background of this style of proof and notation.
In this answer $\;x\;$ is real and $\;n\;$ is integer.$%
\require{begingroup}
\begingroup
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\Ref}[1]{\text{(#1)}}
\newcommand{\true}{\text{true}}
\newcommand{\ceil}[1]{\left\lceil #1 \right\rceil}
%$
We calculate which $\;x\;$ are in $\;\bigcup_{n \ge 2}S_n\;$:
$$\calc
x \in \bigcup_{n \ge 2}\left\{x \mid \tfrac 1 n \le x \lt 1 + \tfrac 1 n \right\}
\op\equiv\hint{expand definition of $\;\bigcup\;$}
\langle \exists n :: n \ge 2 \;\land\; \tfrac 1 n \le x \lt 1 + \tfrac 1 n \rangle
\op\equiv\hint{multiply by $\;n\;$ -- work towards isolating $\;n\;$}
\langle \exists n :: n \ge 2 \;\land\; 1 \le nx \;\land\; n(x-1) \lt 1 \rangle
\op\equiv\hints{divide middle inequality by $\;x\;$, which must be $\;{}>0\;$}\hint{-- further isolating $\;n\;$}
x \gt 0 \;\land\; \langle \exists n :: n \ge 2 \;\land\; n \ge \tfrac 1 x \;\land\; n(x-1) \lt 1 \rangle
\tag{*}
\endcalc$$
Now, to isolate $\;n\;$ in the last inequality, we need to divide by $\;x-1\;$, which can be positive, negative, or zero, so a case split seems unavoidable.
Case $\;x<1\;$.
$$\calc
\Ref{*}
\op\equiv\hint{divide by $\;x-1\;$, which is negative}
x \gt 0 \;\land\; \langle \exists n :: n \ge 2 \;\land\; n \ge \tfrac 1 x \;\land\; n \gt \tfrac 1 {x-1} \rangle
\op\equiv\hint{choose a large enough $\;n\;$}
x \gt 0
\endcalc$$
Case $\;x=1\;$.
$$\calc
\Ref{*}
\op\equiv\hint{substitute $\;x:=1\;$}
1 \gt 0 \;\land\; \langle \exists n :: n \ge 2 \;\land\; n \ge 1 \;\land\; 0 \lt 1 \rangle
\op\equiv\hint{simplify; choose a large enough $\;n\;$}
\true
\endcalc$$
Case $\;x>1\;$.
$$\calc
\Ref{*}
\op\equiv\hints{use $\;2 \gt \tfrac 1 x\;$ to remove $\;n \ge \tfrac 1 x\;$;}\hint{divide by $\;x-1\;$, which is positive}
x \gt 0 \;\land\; \langle \exists n :: n \ge 2 \;\land\; n \lt \tfrac 1 {x-1} \rangle
\op\equiv\hints{apply $\Ref{0}$ below -- to make the rightmost}\hint{expression integer, for the next step}
x \gt 0 \;\land\; \langle \exists n :: 2 \le n \lt \ceil{\tfrac 1 {x-1}} \rangle
\op{\equiv \tag{!}}\hint{arithmetic: simplify using total order on integers}
x \gt 0 \;\land\; 2 \lt \ceil{\tfrac 1 {x-1}}
\op\equiv\hint{apply $\Ref{0}$ again -- to simplify}
x \gt 0 \;\land\; 2 \lt \tfrac 1 {x-1}
\op\equiv\hint{arithmetic using $\;x>1\;$}
0 \lt x \lt 1.5
\endcalc$$
Combining the above, we get
$$\calc
\Ref{*}
\op\equiv\hint{combine the above cases}
0 < x < 1 \;\lor\; x = 1 \;\lor\; 1 < x < 1.5
\op\equiv\hint{}
x \in (0,1.5)
\endcalc$$
which completes the proof.
Here we used the 'ceiling' or 'round up' notation $\;\ceil\ldots\;$, and we used
$$
\tag{0}
m < y \;\equiv\; m < \ceil y
$$
for any integer $\;m\;$ and real $\;y\;$, to convert between integers and reals. That is what allowed the key step $\Ref{!}$.
$%
\endgroup
%$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding complex roots of fourth degree polynomial $z^4 + 8z^3 + 16z^2 + 9$ I have the equation: $$z^4 + 8z^3 + 16z^2 + 9 = 0$$
I need to find all the complex solutions and I've got no clue how to approach it. I've tried factoring but nothing came out of it. I'm still very new to the world of complex numbers so I'll appreciate any help.
| If you substitute $$x=z+2$$ the equation turns into $$x^4-8x^2+25=0$$ which can be solved by a further substitution $y=x^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726195",
"timestamp": "2023-03-29T00:00:00",
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$\mathcal{O}_K$ UFD $\iff h_K=1$ How can we prove that, if $K$ is a number field, then his integer ring $\mathcal{O}_K$ is an unique factorization domain if and only if the class number of $K$ is 1?
| Consider these:
*
*The ideals of $\mathcal{O}_K$ have unique factorization into prime ideals.
*The class number of $K$ is $1$ iff every ideal of $\mathcal{O}_K$ is principal.
| {
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Find a 90% confidence interval for the population variance The variability in the amount of impurities present in a batch of chemicals used for a particular process depends on the length of time that the process is in operation. The standard deviation for the traditional process is 1.24. A new process has been developed to reduce the variability of impurities. Suppose a random sample of operation times is drawn under the new process. The results are given below:
$5.02$, $5.65$, $4.83$, $6.01$, $5.61$, $5.25$, $5.36$, $4.84$, $5.27$, $5.38$, $5.03$, $5.14$, $5.37$, $5.67$, $4.68$, $5.25$
(a) Find a 90% confidence interval for the population variance. What assumption do you need to make?
(b) Since the purpose of the research is to reduce the variability, the smaller the SD the better. Hence only the upper bound SD is of interest. Find a 90% confidence upper bound for the population standard deviation.
(c) Make a statistic inference whether the new process reduces the variability of impurities. Explain.
What I have tried so far:
(a) For the first part, I started by computing the point estimator, where I ended up getting $S = 0.35234$. From there, I used the pivotal quantity $\frac{(n-1)S^2}{\sigma^2}$~$\chi^2(n-1)$. In order to computer the confidence interval, I did the following: $$P(\chi_{0.95}^2 \le \frac{(n-1)S^2}{\sigma^2} \le \chi_{0.05}^2) = P(\frac{1}{\chi_{0.95}^2} \ge \frac{\sigma^2}{(n-1)S^2} \ge \frac{1}{\chi_{0.05}^2}) = P(\frac{(n-1)S^2}{\chi_{0.95}^2} \ge \sigma^2 \ge \frac{(n-1)S^2}{\chi_{0.05}^2}) $$
So the derived confidence interval should come out to be: $(\frac{(n-1)S^2}{\chi_{0.05}^2},\frac{(n-1)S^2}{\chi_{0.95}^2})$ where $n=16$, $S^2 = 0.12414$, $\chi_{0.05}^2 = 24.9958$, and $\chi_{0.95}^2 = 7.2609$. Plugging everything in, I ended up with $(0.074497, 0.25645)$. Here, I believe we have to assume that this is a random sample that follows a normal distribution.
(b) Here I wasn't too sure how to set this up. I thought that if I want the upper limit, I would want to set it up as follows: $$P(\chi_{0.9}^2 \le \frac{(n-1)S^2}{\sigma^2})$$ Again, solving this inequality for $\sigma$ (not $\sigma^2$ since we want SD, not variance), I get $$P(\sigma \le \sqrt{\frac{(n-1)S^2}{\chi_{0.9}^2}})$$ Using the same n and $S^2$, and using $\chi_{0.9}^2=8.54675$, I plug everything in to get $P(\sigma \le 0.46677)$
I'm not sure if I did all of these parts right, but any help would be appreciated. Thanks in advance.
| Seems to me you didn't quite finish. I believe you want a 2-sided confidence interval.
If $L$ and $U$ cut 5% from lower and upper tails of $\mathsf{Chisq}(n-1),$ respectively, then from your first displayed equation a 90% CI for $\sigma^2$ is of the form $\left((n-1)S^2/U,\, (n-1)S^2/L\right).$
| {
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Find value of $t$ where slope of parametrically-defined curve $=4$ using multivariable calculus A recent problem I encountered gave me curve $C$ defined by the parametric equations $x(t)=2t^{2}+t-1$ and $y(t)=t^{2}-3t+1$, and asked me to find point $t$ where the slope of the tangent line to $C$ would equal $4$. Obviously, this is trivial using the Single-Variable equation
$$\frac{\text{d}y}{\text{d}x}=\frac{\text{d}y/\text{d}t}{\text{d}x/\text{d}t}=4,$$
But I'm interested in interpreting $C$ as a literal curve using Multivariable Calculus. I first tried defining the vector-valued function
$$\mathbf{r} = <x(t),y(t)> = <2t^{2}+t-1,t^{2}-3t+1>,$$
but I quickly ran into problems regarding what would constitute the slope of a vector-valued function. One attempt I tried was to evaluate $||\mathbf{r'}||$ and match that to $4$ - that is,
$$||\mathbf{r'}||=4,$$
but that quickly went nowhere, as one might imagine.
Further research indicated to me that the best equivalent of a "slope" in Multivariable Calculus would be via a curve's gradient. However, that approach would require to define $t$ in terms of $x$ and $y$ such that an appropriate function $f(x,y)$ can be defined, a task that would be rather messy in the case of $\mathbf{r}$, even in the case of invoking the Quadratic Formula. This also ignores the lack of a guiding vector $\mathbf{u}$ which I would need to take the dot product of with $\nabla\mathbf{f}$ to get a magnitude resembling slope to match with $4$, or more specifically
$$\nabla \mathbf{f}\cdot\mathbf{u}.$$
My question is thus what would be the best way to approach the problem using Multivariable Calculus, either via expressing $C$ as a vector-valued function or as a function of 2 variables.
| Hint: you will get the equation: $$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2t-3}{4t+1}=4$$. Solve this for $t$!
| {
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Show that this ring has no identity.
Let $R=\left \{ g:\Bbb{R}\to \Bbb{R} \mid g \text{ is continuous and } g(1)=0 \right \}$ be a ring. Show that $R$ has no identity.
The answer says there does not exist a function $h(x)\in R$ such that $h(x)=1$, which I don't understand why, since the only condition in $R$ is $g(1)=0$. Please help me understand what I am missing.
| *
*The most obvious choice of identity function is the constant function $h(x) = 1$. After all, multiplying $h(x)$ pointwise by another function $g(x)$ results in $g(x)$ again.
*Unfortunately, $h(x)$ doesn't belong to the ring $R$ because $h(1)=1\neq 0$. It can't be the identity because it isn't in $R$.
*We can try to repair the problem by defining a different identity function $H(x)$ which is equal to 1 at most points, except $H(1)=0$ as required.
*But then by continuity, $H$ is equal to zero at 1, $H$ is equal to 1 somewhere else, and so at some point $p$, $H(p)=\frac{1}{2}$ (an intermediate value). This is a disaster— take $g(x) = |x-1|$, for example: At that point $p$, $$g(p)H(p) = \frac{1}{2}g(p) \neq g(p).$$ So no such identity function $H(x)$ exists.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to show that the set $Z(G)=\{z \in G : \forall g \in G, z * g = g * z\}$ is a subgroup of $(G, *)$? I have shown that the neutral element is in $Z(G)$. I have also shown that the law is closed in $Z(G)$.
However, I'm not sure how to prove that $\forall x \in Z(G), \exists x' \in Z(G)$ such that $x * x' = x' * x = e$, where $e$ is the neutral element.
I would like to post what I've tried so far, but everything I think of immediately leads to a dead end. How do I go about solving this? Also, we don't know if $G$ is a commutative group. Thank you.
| If $x\in Z(G)$, there is a $x'\in G$ such that $xx'=x'x=e$. So, the problem is to prove that $x'\in Z(G)$. Take $g\in G$. Now, take $g'\in G$ such that $gg'=g'g=e$. Then $xg'=g'x$ (since $x\in Z(G)$), but it follows from this that $gx'=x'g$, as we wanted to prove.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Perigee, Apogee Upper Bounds from Belaga and Mignotte I am currently reading the work of Belaga on upper bounds on minimal cyclic iterates in the $3x+d$ problem.
In the paper, the author gives an upper bound on the perigee as
$$
dk^{C_2}
$$
where $k$ is the number of odd elements in the cycle, and $C_2$ is an effectively computable constant not exceeding $32$ (as per Corollary 2 in the cited paper of Laurent et al).
Later in the introduction, Belaga mentions an upper bound for the apogee (maximal element)
$$dk^C (3/2)^k $$
(the author and Mignotte derive this upper bound in another paper )
Question: Does 32 still apply as an upper bound for the effectively computable constant C (when bounding the apogee)?
The author writes that he corresponded with another author in the derivation of this constant (for the perigee).
| One can apply the results of Rhin (as provided by Lemma 12 in the work of Simons and De Weger) to derive sharp constants.
Assume $k+l>k$. Lemma 12 in Simons/De Weger demonstrates the inequality
$$ (k+l)\log 2 - k\log 3 > e^{-13.3(0.46507)}k^{-13.3}.$$ This inequality provides means for deriving a lower bound on the denominator $2^{k+l}-3^{k}$ of a periodic orbit element; the argument in the abovementioned paper of Belaga/Mignotte demonstrates how this lower bound can be applied to derive an upper bound on the maximal iterate element.
| {
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If $p>0$, then $ \lim_{n\to\infty}\frac{1}{n^p}=0$ using squeeze theorem for sequences.
If $p>0$, then $ \lim_{n\to\infty}\frac{1}{n^p}=0\;.$
Rudin suggests in his Principle of Mathematical Analysis to take $$n> (\frac{1}{\epsilon})^\frac{1}{p}$$ using the Archimedean property of the real number system.
This is under the assumption that we will compute the limit of the sequence based on the fact: If $\ 0 \leq x_n \leq s_n$ for $\ n \geq N$, where N is some fixed number, and if $\ s_n \rightarrow 0$, then $\ x_n \rightarrow 0.$
I don't really understand this proof, but I could try it a different way:
Letting $\ x_n = \frac{1}{n^p}$ and taking $\ s_n$ to be $\ \frac{1}{n}$, then we know that $\ x_n \leq s_n $ because $\ p>0 $. But since $ s_n = \frac{1}{n}$ goes to $0$ as $n$ approaches infinity, we know from $0 \leq x_n \leq s_n$ that $ x_n = \frac{1}{n^p}$ will also go to $0$ as $n$ approaches infinity.
Is this a valid approach?
| What you mentioned is Theorem 3.20 in Rudin's book (page 57).
If one takes $n>(1/\varepsilon)^{1/p}$, then it follows that $n^p>1/\varepsilon$ (because $p>0$) and thus $\displaystyle\frac{1}{n^p}<\varepsilon$. In particular, this argument shows that if one takes a positive integer $N>(1/\varepsilon)^{1/p}$, existence given by the archimedean property, then for any integer $n>N$, one has
$$
n^p>N^p>1/\varepsilon,
$$
equivalently,
$$
\left|\frac{1}{n^p}-0\right|<\varepsilon\;.
$$
This implies by definition of limits that $\displaystyle\lim_{n\to\infty}\frac{1}{n^p}=0$.
Rudin indeed says before the theorem that the squeeze theorem will be used:
But should not take his words too literally ("The proofs will all be ..."). If you take look at the (one line) proof of (e), all he says is "Take $\alpha=0$ in (d)", which is not quite using the squeeze theorem per se.
| {
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Is the set of all Irrational Numbers a ring or a field? I would really appreciate a proof of either one.
I think it should be a field as it satisfies the multiplicative and additive identities and is commutative.
| A ring requires a valid operation (addition) with an inverse (subtraction) and an identity or zero satisfying $x+0=x$ for all $x$.
Likewise a field requires a further operation (multiplication) which is distributive over addition (i.e. $a(b+c)=ab+ac$) and which also has an identity or unit satisfying $1\times x=x$ for all $x$, also having a unique inverse ($\frac{1}{x}$) for every $x$ satisfying $\frac{1}{x}\times x=1$
The other answers here all appear to assume standard binary functions of addition, multiplication etc.
But the irrational numbers $\Bbb{I}$ in themselves are only really become well-defined as an algebraic object once the operations of addition and multiplication are defined.
The irrational numbers are uncountable set and the complex numbers endowed with with standard addition and multiplication are a field.
Therefore a bijection $f:\Bbb{I}\to\Bbb{C}$ can be defined such that $f^{-1}(f(c_1)\cdot f(c_2))$ is a valid addition and $f^{-1}(f(c_1)\circ f(c_2))$ is a valid multiplication in which case $f^{-1}(0)$ is your additive identity and $f^{-1}(1)$ your multiplicative identity.
Once you have done so, $(\Bbb{I},\cdot,\circ)$ is a field and a ring. So in actual fact you are right on both counts, although perhaps not in the sense you expected.
| {
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Number of roots of the equation $x\sin x-1=0$ for $x\in [0,2\pi]$ Number of roots of the equation $x \sin x-1=0$ for $x\in [0,2\pi]$
My attempt is using the bisection method
where I initially took $a=0$, $b=\frac{\pi}{2}$
as $f(a)\cdot f(c)<0$
We can proceed along the lines
we get one root $\in[0,\frac{\pi}{2}]$
Similarly we get another root $\in[\frac{\pi}{2},\pi]$
as $f(a)\cdot f(c)<0$ again
but no roots $\in[\pi,2\pi]$
as $f(a)\cdot f(c) \not< 0$
Hence we can concude we have two roots $\in[0,2\pi]$
Please tell me if I am on the correct lines. Ideas,solutions are appreciated.
And if possible tell me tricks to solve such types of transcendental equations.
The curve of xsinx =0 is given above. See that the ampltude of the curve goes on increasing as $(\frac{\pi}{2},\frac{5\pi}{2}...)$
This proves that if $x \in [0,2\pi]$, the function $xsinx$ can attain a value $=1$ only at two instances i.e on the +ve half of the first wave. Hence we are done .
This is a lot handwavy still gives us a correct answer I guess.
| Here is the plot of f(x) = $x \sin(x) - 1$ for $0\le x \le 2\pi $.
If $f(a)f(c)\lt0$ there must be at least one root between $a$ and $c$ but there could be more!
Also if $f(a)f(c)\gt0$, it does not mean that there is no root between $a$ and $c$. In fact, you could have infinite number of roots there. There is no simple way to tell the number roots in general case.
In this particular case it is obvious that $x\sin(x) \le 0$ for $\pi \le x \le 2\pi$ so there's no root there.
For $0 \le x \le \pi$, the function $x\sin(x)$ goes from zero to zero and passes through $\pi/2$ for $x=\pi/2$. It means that the graph of the functon $x\sin(x)$ must cross line $y=1$ at least twice so you have at least two roots in this region.
| {
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How to find the value of $x$ that satisfies $3x=4$ in $\mathbb Z/5\mathbb Z$? Let $\mathbb Z_5 = \mathbb Z/5\mathbb Z$.
The value of $x$ which satisfies the equation $3x = 4\bmod 5$ is...?
The answer is $3$. I understand why the answer is $3$, but not how it was derived. Is there an equation or process I can use that will give me the correct answer no matter how large the numbers in the equation?
| Consider equations of the form $m\cdot x = n\bmod p$, where $p$ is a prime number.
The other answer is a correct way to derive the solution, but it takes as many as $p$ tries to find it in the worst case. A more efficient solution is the following. First, compute $m^{-1}$, that is, the only solution $y$ to the equation $m\cdot y = 1\bmod p$. This can be done using the extended euclidean algorithm. Then, let $x:= m^{-1}\cdot n$. One can then check that $m\cdot x = mm^{-1}n \bmod p = n\bmod p$.
In case $p$ is not a prime number, one can also use this method in the case that $m$ and $p$ are coprime (which guarantees the existence of $m^{-1}$). Suppose now that $p$ and $m$ share a factor, and call it $q$. If $n$ is not divisible by $q$, then there is no solution (you can prove this using arithmetic modulo $q$). If $q$ also divides $n$, then the equation reduces to $\frac{m}{q}\cdot x = \frac{n}{q}\bmod \frac{p}{q}$. You can now continue this process, until you find a solution.
| {
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Parametric equation of a curve: a line in a circle transform to a curve in an ellipse Giving the circle with a line segment inside, if the circle was stretched into an ellipse, what is the parametric equation of the parabolic curve (I assume) transformed from the line segment?
circle to ellipse
I want the curve as in shown in the pic. Scaling axis only give the line segment with a new slope. I could draw the curve by interpolated the line according to the displacement field inside the circle, but I don't know the parametric equation of the curve. Could anyone give me some hints how to determine the equation in trigonometric way?
| How do you define your stretch transformation? In any transformation a relation between variables before and after transformation is defined.
Setting a variable in one system to a constant we can map the curve in the other system. Like eg polar to cartesian coordinates $ r=a \rightarrow x^2+y^2= a^2. $
I could not figure out transfomation relations in your question. Are any lengths conserved?
Setting aside a full approach if we proceed just with the relation you gave that automatically defines and additionally could take care of its relation to the circle:
$$ x = a \cos \theta + c,$$
by converting to polar coordinates we get directly (no parametrization) the curve Conchoid of Nichomedes:
$$ r \cos \theta = a \cos \theta + c \,; \quad r= a + c \cdot \sec \theta, $$
where the unit circle is added for comparision. For the curve graph $ (a=1, c= 0.3 ) .$
As stated above the strectch definition has still to be looked into.
| {
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Function’s graph - a homework problem Does there exist an $f(x)$ function $R\to R$, such that every line parallel to the $x$-axis meets the function’s graph
a) exactly three times?
b) an even number of times?
For b), I have several examples, which almost work, but there is a line which is tangent to the curve, so it does not work. Please help!
| Hint: For (a), consider $f(x)$ on a single interval of $\mathbb{R}$. You could try partitioning $f(x)$ into $3$ sections over this interval. You could then repeat this in every other interval but using a different range of $y$-values. You could use a different number of partitions for (b). An example function is shown in the figure below:
This is because you want a function where every $y$ value corresponds to exactly $3$ $x$-values. But this can't be a continuous function. To see why this is the case, look at the horizontal purple line in the figure below. In order to have $f(x)$ (in blue) cross the purple line exactly $3$ times, $f(x)$ would need to change direction twice (i.e. have $2$ points of inflection, if $f(x)$ is differentiable). Hence, for $y$-values more extreme than the local maxima of $f(x)$, the function would only reach each $y$ value once.
Given that we know $f(x)$ must be discontinuous to exist at all, perhaps you could try splitting the $x$-axis into $3$ sections, and defining a part of your function on each interval, as in the figure below. Thanks to @polfosol for the inspiration for this. However, this may not work either, as $f(x)$ would need to have a value at each discontinuity (i.e. the orange and green points in the figure below). And the value at the discontinuity can't be in $\mathbb{R}$ since every value's already been previously used by the red curves!
Maybe you could also try mapping the decimal expansion of each real number to a number, in such a way that each output corresponds to exactly $3$ inputs. However, I haven't figured out whether this could be done.
| {
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Why this is wrong let, n be a positive integer.Then,${C_r}=$$ {n} \choose {r}$.Now evaluate $ {C_0}- {C_1}/2+ {C_2}/3+.....+ (-1)^n {C_n}/(n+1)$
I expand $(1-x)^n$ and integrating both side and putting $x=1$ the required series comes.But it gets 0.But answer is $1/n+1$
| Another way:
$$\dfrac{\binom nr}{r+1}=\cdots\dfrac{\binom{n+1}{r+1}}{n+1}$$
$$\sum_{r=0}^n\dfrac{(-1)^r\binom nr}{r+1}=\dfrac1{n+1}\sum_{r=0}^n\binom{n+1}{r+1}$$
Now $$\sum_{r=0}^n(-1)^r\binom{n+1}{r+1}=\binom{n+1}0-(1-1)^{n+1}$$
| {
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Number of $6$-digit numbers made up of the digits $1$, $2$, and $3$ with no digit occurring $3$ or more times consecutively? Find the number of 6-digit numbers made up of the digits $1$, $2$, and $3$ that have no digit occur three or more times consecutively. (For example, $123123$ would count, but $123111$ would not.)
We know the total number of possibilities without the occurrence restriction is $3^6 = 729$.
In order to eliminate the possibilities that can't be used, I set a block of 3 numbers as my 3 consecutive digits block. This block would have 3 different combinations. I then split the number of eliminations into 2 cases: one where the block of 3 is in the front, and one where the block of 3 is in the middle.
For case 1: I computed $3\cdot 24$, with $3$ being the ways to change the block of 3, and $24$ being the number of ways to change the other 3 numbers while not having another block of 3 consecutive.
For case 2: I computed $3^4$, ignoring the restriction of case 1.
Summing the 2 cases, I got $72+ 3\cdot 81 = 315$, I then subtracted from $729$ and got $414$, which turns out to be incorrect. What am I doing wrong?
| There are $3^6 = 729$ possible sequences. From these, we must subtract those in which three consecutive digits are the same.
Observe that a prohibited sequence must begin in one of the first four positions. Let $A_1, A_2, A_3, A_4$ be the set of outcomes in which three consecutive digits beginning in the first, second, third, and fourth positions, respectively, are the same.
The set of outcomes that violate the restriction is $A_1 \cup A_2 \cup A_3 \cup A_4$. By the Inclusion-Exclusion Principle,
\begin{align*}
|A_1 \cup A_2 \cup A_3 \cup A_4| & = |A_1| + |A_2| + |A_3| + |A_4|\\
& \quad - |A_1 \cap A_2| - |A_1 \cap A_3| - |A_1 \cap A_4| - |A_2 \cap A_4| - |A_3 \cap A_4|\\
& \quad + |A_1 \cap A_2 \cap A_3| + |A_1 \cap A_2 \cap A_4| + |A_1 \cap A_3 \cap A_4| + |A_2 \cap A_3 \cap A_4|\\
& \quad - |A_1 \cap A_2 \cap A_3 \cap A_4|
\end{align*}
$|A_1|$: There are three ways of selecting the number that occupies the first three positions and three ways of selecting the number that occupy each of the last three positions. Hence, $|A_1| = 3^4$. By symmetry,
$$|A_1| = |A_2| = |A_3| = |A_4| = 3^4$$
$|A_1 \cap A_2|$: Since the first three and the second three positions overlap, the first four positions must be filled with the same number. There are three ways to select this number and three ways to fill each of the remaining two positions. Hence, $|A_1 \cap A_2| = 3^3$. By symmetry,
$$|A_1 \cap A_2| = |A_2 \cap A_3| = |A_3 \cap A_4| = 3^3$$
$|A_1 \cap A_3|$: Since the first three and the third three positions overlap, the first five positions must be filled with the same number. There are three ways to select this number and three ways to fill the remaining position. Hence, $|A_1 \cap A_3| = 3^2$. By symmetry,
$$|A_1 \cap A_3| = |A_2 \cap A_4| = 3^2$$
$|A_1 \cap A_4|$: Since the first three and last three positions do not overlap, the first three positions can be filled in three ways, as can the last three. Hence,
$$|A_1 \cap A_4| = 3^2$$
$|A_1 \cap A_2 \cap A_3|$: Since these sets overlap, the first five positions must be filled with the same number. There are three ways to choose this number and three ways to fill the remaining position. Thus, $|A_1 \cap A_3 \cap A_3| = 3^2$. By symmetry,
$$|A_1 \cap A_2 \cap A_3| = |A_2 \cap A_3 \cap A_4| = 3^2$$
$|A_1 \cap A_2 \cap A_4|$: Since the first three positions overlap with the second three and the second three positions overlap with the last three positions, all six positions must be the same. There are three ways to select the number that fills all six positions. Hence, $|A_1 \cap A_2 \cap A_4| = 3$. By symmetry,
$$|A_1 \cap A_2 \cap A_4| = |A_1 \cap A_3 \cap A_4| = 3$$
$|A_1 \cap A_2 \cap A_3 \cap A_4|$: All six positions must be filled with the same number. Since there are three ways to select the number,
$$|A_1 \cap A_2 \cap A_3 \cap A_4| = 3$$
By the Inclusion-Exclusion Principle,
\begin{align*}
|A_1 \cup A_2 \cup A_3 \cup A_4| & = 3^4 + 3^4 + 3^4 + 3^4\\
& \quad - 3^3 - 3^2 - 3^2 - 3^3 - 3^2 - 3^3\\
& \quad + 3^2 + 3 + 3 + 3^2\\
& \quad - 3\\
& = 237
\end{align*}
Hence, there are $729 - 237 = 492$ admissible sequences.
| {
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Evaluating $\prod^{100}_{k=1}\left[1+2\cos \frac{2\pi \cdot 3^k}{3^{100}+1}\right]$
Evaluate$$\prod^{100}_{k=1}\left[1+2\cos \frac{2\pi \cdot 3^k}{3^{100}+1}\right]$$
My attempt:
$$1+2\cos 2\theta= 1+2(1-2\sin^2\theta)=3-4\sin^2\theta$$
$$=\frac{3\sin \theta-4\sin^3\theta}{\sin \theta}=\frac{\sin 3\theta}{\sin \theta}$$
I did not understand how to solve after that. Help required.
| Let $$z=\cos\bigg(\frac{2\pi}{3^n+1}\bigg)+i\sin\bigg(\frac{2\pi}{3^n+1}\bigg)$$
Then $z^{3^n+1}=1$ and also $\displaystyle 2\cos \bigg(\frac{2\pi\cdot 3^k}{3^n+1}\bigg)=z^{3^k}+\frac{1}{z^{3^k}}$
Write $$\prod^{n}_{k=1}\bigg[1+2\cos\bigg(\frac{2\pi\cdot 3^k}{3^n+1}\bigg)\bigg]$$
$$=\bigg(1+z^3+\frac{1}{z^3}\bigg)\bigg(1+z^9+\frac{1}{z^9}\bigg)\cdots \cdots \bigg(1+z^{3n}+\frac{1}{z^{3n}}\bigg)$$
$$=\frac{(1+z^3+z^6)(1+z^9+z^{18})\cdots \cdots (1+z^{3^n}+(z^{3^n})^2)}{z^{3+9+\cdots \cdots +3^n}}$$
Multiply both Nr and Dr by $(1-z^3)$
$$=\frac{1-z^{3^{n+1}}}{(1-z^3)\cdot z^{3\frac{(3^n-1)}{2}}}= \frac{1-z^{-3}}{-(1-z^3)\cdot z^{-3}}=1.$$
$\text{Simplification}:\;\; $ From $z^{3n+1}=1\Rightarrow z^{3n}=z^{-1}$
And $$z^{3\frac{(3^n-1)}{2}}=z^{-3}\cdot z^{3\frac{(3^+1)}{2}}=z^{-3}\cdot \bigg(z^{\frac{(3^+1)}{2}}\bigg)^3=-z^3$$
| {
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Integral of $ \int \frac {\tanh(x) dx}{\tanh(x)+\operatorname{sech}(x) }$ My attempt at solution:
Also I got a question, is there a way this can be solved without using hyperbolic tangential half angle substitution? Because I don't get how you can treat hyperbolic functions as if they were trigonometric and deducing sinx and cosx from a right triangle
| The integral can also be seen as the following:
\begin{align}
\int \frac{\tanh(x) \, dx}{\tanh(x) + sech(x)} &= \int \frac{\sinh(x) \, dx}{\sinh(x) + 1} = \int \left( 1 - \frac{1}{1 + \sinh(x)} \right) \, dx \\
&= x + \sqrt{2} \, \tanh^{-1}\left( \frac{1}{\sqrt{2}} \, \left(1 - \tanh\left(\frac{x}{2}\right) \right) \right)
\end{align}
| {
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Why is the value of the series representation for $\arctan$ at $ x = 1$ necessarily $\arctan(1)$ Exercise 6.6.1. The derivation in Example 6.6.1 shows the Taylor series for $\arctan(x)$ is valid for all $x \in (−1,1)$. Notice, however, that the series also converges when $x = 1$. Assuming that $\arctan(x)$ is continuous, explain why the value of the series at $x = 1$ must necessarily be $\arctan(1)$. What interesting identity do we get in this case?
This is a question from Abbott's Understanding Analysis. My question is what exactly constitutes a proof that the series equals the function at a point.
| Proof. Since $arctan(x)$ is continuous, for all $\epsilon > 0 $ there exists some $\delta > 0 $ such that $|1-x| < \delta $ implies that $|arctan(1) - arctan(x)| < \epsilon $. We also know that $$arctan(x) = x - \frac 1 3 x^3 + \frac 1 5 x^5 - \frac 1 7 x^7 + \cdots $$ when $ x \in (-1, 1)$. Hence $|1-x| < \delta $ implies that $$|arctan(1) - ( x - \frac 1 3 x^3 + \frac 1 5 x^5 - \frac 1 7 x^7 + \cdots )| < \epsilon $$ for all $\epsilon>0$. Hence $$arctan(1) = 1 - \frac 1 3 1^3 + \frac 1 5 1^5 - \frac 1 7 1^7 + \cdots$$ as desired.
| {
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Why is the integral defined as the limit of the sum $\int_a^b f(x) dx = \lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\Delta x$? I am failing to understand why the integral is defined as:
$$\int_a^b f(x) dx = \lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\Delta x$$
instead of:
$$\int_a^b f(x)dx=\sum_{i=1}^\infty f(x_i^*)\Delta x$$
Is the former just popular preference or is there something I am not conceptually understanding here?
| Actually, the definition is:
Let $\mathcal{P}=\{a, x_1, \ldots, x_k, b\}$ be a partition of $[a, b]$ and denote $\Delta \mathcal{P} = \max_i |x_i-x_{i+1}|$. Then we say a bounded function $f$ is Riemann integrable provided for any $\varepsilon>0$ there exists $\delta>0$ such that if $\Delta \mathcal{P}<\delta$ implies
\begin{align}
\left|\sum_{\mathcal{P}} \left\{M_i-m_i \right\}(x_{i+1}-x_i)\right|<\varepsilon
\end{align}
where
\begin{align}
M_i:=\sup_{t \in[x_i, x_{i+1}]}f(t) \ \ \text{ and } \ \ m_i:=\inf_{t \in[x_i, x_{i+1}]}f(t).
\end{align}
The partition is finite (just like how partial sum is finite).
Once you know that $f$ is Riemann integrable, then you can define the integral of $f$ to be
\begin{align}
\int^b_a f(t)\ dt:= \lim_{\Delta\mathcal{P}\rightarrow 0}\sum_{\mathcal{P}} M_i (x_{i+1}-x_i).
\end{align}
Once you know $f$ is Riemann integrable, then you can specialize a nested sequence of partition $\mathcal{P}_n \subset \mathcal{P}_{n+1}$ with $\Delta \mathcal{P}_n \geq \Delta \mathcal{P}_{n+1}\rightarrow 0$ so that
\begin{align}
\int^b_a f(t)\ dt = \lim_{n\rightarrow \infty} \sum^n_{i=1} f(x^\ast_i)\Delta x_i
\end{align}
is actually meaningful.
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Union of two compact totally disconnected sets in $\mathbb{R}$ is totally disconnected I'm so stuck with the next problem. First, the definition that I have are the next:
$X$ is totally disconnected if for all $x\in X$ we have that $C_x=\{x\}$ where $C_x$ is the connected component.
Let $A,B\subseteq\mathbb{R}$ be a compact and totally disconnected sets. Prove that $A\cup B$ is also totally disconnected.
We have two cases:
1) If $A\cap B=\emptyset$ then, take $x\in A\cup B$ and $C_x$ the connected component of $x$. Because $\mathbb{R}$ is normal and $A,B$ are two disjoint closed sets (they are compacts) then there exists $U,V\subseteq\mathbb{R}$ disjoint open sets such that $A\subseteq U$ and $B\subseteq V$. Therefore, without loss of generality, because $C_x$ is connected, $C_x\subseteq U$ and thus $C_x\subseteq A$. We conclude that $C_x=\{x\}$.
2) $A\cap B\neq\emptyset$. Here is where I'm stuck. Take $x\in A\cup B$ and suppose by contradiction that $|C_x|\geq 2$. Then, $C_x\not\subseteq A$ and $C_x\not\subseteq B$ (because the connected sets in $A$ and $B$ have only one point) but $C_x\cap A\neq\emptyset$ and $C_x\cap B\neq\emptyset$. I don't know how can I conclude the proof. Any hint? I really appreciate any help you can provide.
| First of all recall that a subset of $\mathbb{R}$ is connected if and only if it is an interval. The same holds for intervals as well: a subset of an interval is connected if and only if it is again an interval. Here by interval I understand a subset $I\subseteq\mathbb{R}$ such that if $a,b\in I$ and $a<c<b$ for some $c\in \mathbb{R}$ then $c\in I$.
Assume that $A\cup B$ is not totally disconnected. Then $[a,b]\subseteq A\cup B$ for some $a,b\in A\cup B$, $a<b$. Let $I=[a,b]$.
Since both $A,B$ are compact (actually closed in $\mathbb{R}$ is enough) then $I\cap A$ and $I\cap B$ are closed. Both are nonempty and their union is $I$ and thus we can rephrase the problem:
Show that a closed interval $I=[a,b]$, $a<b$ cannot be written as a union of two non-empty, closed and totally disconnected subsets.
Let $I=A\cup B$. Both subsets have to be proper otherwise they wouldn't be totally disconnected. Since $A$ is closed then $I\backslash A$ is open in $I$. Thus $I\backslash A\subseteq B$ and so $B$ is not totally disconnected because it contains an open subset of $I$ (which contains an interval by definition). Contradiction.
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The image of $\emptyset$ under $f$ is empty. Let A= $\{ \emptyset,\{1\},\{2,\},...,\{n\} \}$. Define $f\colon A \to \mathbb{N}$ such that $f(\emptyset)=0$ and $f(\{n\})=n$.
Is this a counterexample that the image of the $\emptyset$ under an arbitrary function $f$ is empty?
| You are defining a function on a set $X$ that has $\emptyset$ as an element. So $f(\emptyset)$ can be anything you like.
But you're probably thinking of the notion of "image of a subspace under $f$".
I (and many texts/papers) always separate these notions notationwise, using square brackets so $f[\emptyset] = \emptyset$ which is indeed always true, and in general $$f[A]= \{y : \exists x \in A: f(x) = y\}$$ is used for the set of $f$-images of elements of $A$. Clearly, if $A$ has no elements, there are also no $f$-images. Some texts write $f(A)$ for $f[A]$ but this can get confusing, as witnessed by your question.
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Angle trisection of $90^o$ Read on page #7 of article here, that angle of $90^o$ can be trisected. I went through this youtube video here, and here; and denote these two videos (denoting two separate methods) by (a), (b) respectively. These use elementary methods used in school days, but were never explained for the reason.
Anyway, I want to understand the logic behind both (a), (b) as it is a valid trisection.
In (a) there is an arc equal to the length of the radius, let $|r|$. Now, I hope that it is an approximation as there should be (as per me) no way to get the $\frac{2}{3}: \frac{1}{3}$ on the $90^o$ arc. It is only the chord joining the two ends (let, $yx$ with $y$ the coordinate on the $y$-axis, and $x$ the coordinate on the $x$-axis) that can be hoped to be approximately divided by the arc from point $y$ or $x$. Either end's arc will cut the hypotenuse $yx$ in two parts with ratios $\frac{1}{\sqrt{2}}= 0.707, 1-0.707=0.293$ in opposite sides from point $y,x$ respectively. So, if I am not wrong, it is an approximate approach and hence not mathematically or even geometrically valid.
Approach shown in (b) is even difficult to understand, and would request some help, as it is using a chord's division (named $DE$) in two parts (at midpoint $F$) by the angle bisector of $90^o$ to derive further angle bisections on both sides of the midpoint $F$. I am unable to comprehend even the mechanism in this approach, as the average of $45^o $ and $90^o$ is $67.5^o$, while the average of $0^o$ and $45^o$ is $22.5^o$.
Also, both of the above methods are faulty, then please give some link for correct way.
Edit Unable to understand, and hope that the approach (a) is valid, as a similar approach is used here, in Fig. 3.13. But, am unable to understand that too, but being a credible reference source, can infer (a) is at least a valid one.
Edit 2 I am posting an image of the reference in the 'Edit' made earlier, for a different, but linked question. It is given on the earlier page (#53) that on bisecting the smaller angle $\widehat {AON}$ obtained by bisecting, get the trisection. By this logic, the approach (b) is also valid. But, how can it be understood, is not clear, as double bisection to lead to $\frac{1}{4}$th of the original angle.
Edit 3 Regarding bisection leading to trisection, have found a suitable basic geometrical technique at: http://www.lamath.org/journal/Vol1/trisection_using_bisection.pdf, on pg. #3,4, by using the two unequal circles' intersection formed from the bisection of the original interval.
The article titled: 'Trisection using Bisection' describes also the nth-section of any interval using unmarked ruler and collapsing compass, in the last section.
|
The line $AB$ is perpendicular to $BC$. The circles have, respectively, centers $B$ and $C$ and radius $BC$. The segments $BD$, $BC$ and $CD$ are equal by construction. Can you prove that the angle $\widehat{CBD}$ is twice the angle $\widehat{DBA}$?
| {
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$f$ convex, $g$ concave and increasing, $\int_0^1 f = \int_0^1 g$, then $\int_0^1(f)^2 \geq \int_0^1(g)^2$
Let $f,g:[0,1] \to [0, \infty)$ be two continuous functions such that $$f(0) = g(0) = 0,$$ $f$ is convex, $g$ is concave and increasing and $$\displaystyle \int_0^1f(x)dx = \int_0^1g(x)dx.$$ Prove that $$ \displaystyle \int_0^1\left(f(x) \right)^2dx \geq \int_0^1\left(g(x) \right)^2dx.$$
I don't quite know how to approach the problem.
I thought about using Chebyshev's inequality due to the fact that $g$ is increasing, $F(x) = \int_0^xf(t)dt$ is increasing (since $f$ is convex) and $G(x) = \int_0^xg(t)dt$ is increasing (since $g$ is increasing and $g(0) = 0$), but it didn't help me.
I also tried to obtain something by writing the convexity and concavity point-wise and using the fact that $\displaystyle h(x) = \frac{f(x)}{x}$ is increasing and $p(x) = \displaystyle \frac{g(x)}{x}$ is decreasing, but I got nothing.
| For some $0 < \epsilon < 1$,
$$
\int_\epsilon^1 (f(x)^2-g(x)^2)dx = \int_\epsilon^1 (f(x) + g(x))(f(x)-g(x)) dx
$$
since $f(x)+g(x) \ge 0$, we can apply the intermediate value theorem to get a $\xi \in [\epsilon,1]$, such that
$$
\int_\epsilon^1 (f(x) + g(x))(f(x)-g(x))dx = (f(\xi)+g(\xi))\int_\epsilon^1 (f(x)-g(x))dx \,.
$$
Since $f(\xi)+g(\xi) \ge 0$, it remains to show that
$$
\int_\epsilon^1 (f(x)-g(x))dx \ge 0
$$
$h(x) = (f(x)-g(x))$ is also convex ($-g(x)$ is convex) with ($h(0) = 0$), so $h$ can only have at most 2 roots (except $h \equiv 0$). If there is no other root $x_0 \in(0,1)$, then either $f(x) > g(x)$ or the other way around, and the integrals would not match. So there must be $x_0 \in(0,1)$ such that $h(x) \ge 0$ for all $x \ge x_0$. Therefore $f(x) \ge g(x)$ and furthermore $\int_\epsilon^1(f(x)-g(x))dx \ge 0$, with $\epsilon := x_0$.
| {
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Proof that series converges in probability. Consider $X_{1} \dots X_{n} \dots$ - independent random variables with Cauchy-distribution with scale parameter $\theta_{n}$ and zero location parameter such as : $\sum_{n \ge 1} \theta_{n}$ converges. Prove that $X = \sum_{n \ge 1} X_{n}$ converges in probability.
My attempt : let's try to use Cauchy-criteria
$\forall \epsilon , \delta \exists N $: $\forall n, m \> N $ we have $\operatorname{\mathbb{P}}(| \sum_{n \le k \le m} X_{k}| \ge \epsilon) \le \delta$. Now let's use Markov's inequality $LHS \le \frac{\sum \operatorname{\mathbb{E}(X_{k})}}{\epsilon}$. And I thought that it's converges to zero. But I know there isn't exist a expected value of Cauchy random variable.
Also I thought about : I can prove that sum of Cauchy variables with parameters $\{\theta_{n}\}$ is a random variable with parameter to be equals $\theta = \sum \theta_{n}$. So I have a random variable with Cauchy distribution. But how can it help me ?
Any advices ?
| By Cauchy criterion for convergence in probability it suffices to prove that $$\lim_{n,m\to \infty}P\left(\left|\sum_{k=n+1}^m X_k \right| > \epsilon \right) = 0$$
A simple computation with characteristic functions proves that $\sum_{k=n+1}^m X_k$ follows a Cauchy distribution with location $0$ and scale parameter $\sum_{k=n+1}^m \theta_k$. Since this distribution is symmetric, $$P\left(\left|\sum_{k=n+1}^m X_k \right| > \epsilon \right) = 2P\left(\sum_{k=n+1}^m X_k > \epsilon \right)=2(1-\frac 1\pi \arctan\left(\frac{\epsilon}{\sum_{k=n+1}^m \theta_k}\right)-\frac 12)$$
By the original assumption, $\lim_{n,m\to \infty}\sum_{k=n+1}^m \theta_k = 0$, thus $\lim_{n,m\to \infty}\frac 1\pi \arctan\left(\frac{\epsilon}{\sum_{k=n+1}^m \theta_k}\right) = \frac 12$ and the result follows.
| {
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Compute the limit of the sequence ${\textstyle\sum_{k=1}^n}\frac1{\sqrt{n^2+k}}$ I have to compute the limit of this sequence ${\textstyle\sum_{k=1}^n}\frac1{\sqrt{n^2+k}}$ as $n\rightarrow\infty$. First I was thinking about some Riemann sum and and forced the $n^{2}$ outside the square root but the function was not so pleasant.
| How about squeezing ?
$$\frac{n}{\sqrt{n^2+n}}\leq \sum_{k=1}^n\frac1{\sqrt{n^2+k}}\leq \frac{n}{\sqrt{n^2+1}}$$
The outer terms both go to $1$.
| {
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Graph Theory - proving equivalence
Let $G$ be a graph with order $n ≥ 2$. Prove that $G$ satisfies $(a)$ if and only if it satisfies $(b)$.
$(a)$ for any pair of distinct vertices $u, v$ of $G$, there exists at least one $(u, v)$-path in $G$;
$(b)$ for any partition ${X, Y }$ of $V (G)$ into two non-empty parts $X, Y$ , there exists at least one edge $e = xy$ of $G$ such that $x ∈ X$ and $y ∈ Y$.
This is quite easy to understand intuitively but I'm having trouble trying to prove it, particularly proving $(b) \implies (a)$. Here's my approach to $(a) \implies (b)$:
Suppose $(a)$ is true, then $G$ is a connected graph. Let $X$ and $Y$ be $2$ non-empty partitions of $V(G)$. Suppose there is no such $e=xy$ of $G$ such that $x ∈ X$ and $y ∈ Y$. This means $G$ is disconnected which is a contradiction. So there exists at least one edge $e = xy$ of $G$ such that $x ∈ X$ and $y ∈ Y$.
I'm new to graph theory and proofs as such in general. I would really appreciate some feedback on the attempt above and some help with proving $(b) \implies (a)$. Thanks
| You made use of the word "connected" all too quickly. We have to get hold of this notion first. Call two vertices $u$,$v\in V$ equivalent if there is an edge path in $G$ connecting $u$ and $v$.
Make sure that this is indeed an equivalence relation on the set $V$ of vertices of $G$. How many equivalence classes can there be if (b) holds?
| {
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Let $S$ be a metric space, $ f: S \to \Bbb R$ continuous. Define $Z(f) = \{p \in S : f(p) = 0 \}$. Prove $Z(f)$ is closed. Let $S$ be a metric space, $ f: S \to \Bbb R$ continuous. Define $Z(f) = \{p \in S : f(p) = 0 \}$. Prove $Z(f)$ is closed.
I've come up with a proof... I just would like to know if it is logical enough or it needs to be improved upon somehow. It seems very long for a simple idea.
Below is my proof:
Must show $\Bbb R \setminus Z(f)$ is open.
$\quad \Bbb R \setminus Z(f)= \{x: f(x) \ne 0\} = \{x: f(x) > 0\} \cup \{x: f(x) <0\}$
It's enough to show $\{x: f(x) > 0\}$ is open if $f$ is continuous, because $\{x: f(x) < 0\} = \{x: -f(x) > 0\}$, and $-f$ is continuous if $f$ is continuous.
So, let $x \in \Bbb R$ be such that $f(x) > 0$. There exists $\epsilon > 0$ such that $f(x) > \epsilon$.
Say $\epsilon = \frac{f(x)}{2}$, since $f$ is continuous, there exists $\delta >0$ such that
$\quad |f(x) - f(y)| < \epsilon = \frac{f(x)}{2}$ if $|x-y| < \delta$
then,
$\quad-\frac{f(x)}{2} < f(y) - f(x) < \frac{f(x)}{4}$
so,
$\quad f(y) > f(x) - \frac{f(x)}{2} = \frac{f(x)}{2} > 0$
since $y$ with $|y - x| < \delta$ is arbitrary, $(x - \delta, x + \delta) \subset \{x : f(x) > 0 \}$
Therefore $Z(f)$ is closed.
| Another way is to use sequential characterization, so let $(p_{n})\subseteq Z(f)$ be such that $p_{n}\rightarrow p$ in $S$, then continuity of $f$ gives $f(p_{n})\rightarrow f(p)$. But $f(p_{n})=0$, so $f(p)=0$, this shows that $p\in Z(f)$, we are done.
| {
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Showing the sequence converges $a_{1}=\frac{1}{2}$, $a_{n+1}=\frac{1}{2+a_{n}}$ Showing the sequence converges $a_{1}=\frac{1}{2}$, $a_{n+1}=\frac{1}{2+a_{n}}$.
I already know that if $(a_{n})$ converges then it does to $\sqrt{2}-1$.But i dont't know how to prove that this sequence cenverges.
EDIT
I think that the subsequence $(a_{2n+1})$ is monotonic decreasing and the subsequence $(a_{2n})$ is monotonic increasing.
| Since you already have a reasonable conjecture consider the sequence $(b_n)_{n\geq1}$ defined by
$$a_n=\sqrt{2}-1+b_n,\qquad{\rm resp.,}\qquad b_n=a_n+1-\sqrt{2}\qquad(n\geq1)\ ,$$
and try to prove that $\lim_{n\to\infty} b_n=0$, which should be simpler.
| {
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Problem Understanding a Definition of Order of a Field I need some help regarding some definitons.
I was studying algebraic number theory and I am stuck on this.Can someone explain me what is meant by $F = [\mathcal{O}_k:\mathcal{O}]$ with an easy examples
I was studying the following theorem but didn't get examples
Let $\mathcal{O}$ be an order in $K$. Then $F = [\mathcal{O}_k:\mathcal{O}]$ is finite and
$\mathcal{O} = \mathbb{Z} + F \mathcal{O}_k$.
Moreover, any set $\mathcal{O} = \mathbb{Z} + F \mathcal{O}_k$. with $1 \leq F$ is an order in $K$ such that
$F = [\mathcal{O}_k:\mathcal{O}]$
| Since you mentioned that you're studying algebraic number theory, I'm assuming $K$ is an algebraic number field in your context.
An order $\mathcal{O}$ of a number field $K$ is defined to be a subring of the ring of integers $\mathcal{O}_K$ with $d$ generators over $\mathbb{Z}$, where $d$ is the degree of the number field.
Thus, $F = [\mathcal{O}_K : \mathcal{O}]$ simply means the index of the subring $\mathcal{O}$ in $\mathcal{O}_K$.
For example, if we are working with the number field $K = \mathbb{Q}(\alpha)$ where $\alpha^2 + 1 = 0$. One can show that the ring of integers is $\mathcal{O}_K = \mathbb{Z}[\alpha]$. Choose your favorite integer $z \in \mathbb{Z}$. Note that $\mathcal{O} = \mathbb{Z}[z\alpha] = 1\mathbb{Z} + z\alpha\mathbb{Z}$ is a subring of $\mathcal{O}_K$. Hence, $\mathcal{O}$ is an order of the number field $K$. In this case, $F = [\mathbb{Z}[\alpha] : \mathbb{Z}[z\alpha]] = z$.
You might also want to edit your question to make it more precise.
| {
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Dividing polynomials in $\mathbb{F}_7[x]$ I am trying to divide $4x^4+3x^3+2x^2+x+1$ by $2x^2+x+1$ in $\mathbb{F}_7[x]$.
Normally outside of $\mathbb{F}_7[x]$ I know that the answer would be $2x^2+(1/2)x-(1/4)$ with a remainder of $(3/4)x+(5/4)$. But because this is in $\mathbb{F}_7[x]$ and the coefficients must be in $\mathbb{F}_7$, I am confused on how to deal with the fractions. Can anyone help?
| $\frac{1}{2}=\frac{1×4}{2×4}=\frac{4}{8}=\frac{4}{1}$ as 8 mod 7 is 1.Here you need to make denominator 1 by selecting suitable number.
$\frac{-1}{4}=\frac{6×2}{8}=\frac{6×2}{1}=\frac{5}{1}$. Since $-1=6 mod 7$.
Similarly you can try other fraction mod 7.
| {
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Show that if the integral of the "derivative" of a L^1 function is 0, then the function is constant a.e. I'm working on the following problem:
Let $f\in L^1[a,b]$. Prove that if
$$\lim\limits_{h\to 0}\frac 1h\int_a^b|f(x+h)-f(x)|dx=0,(*)$$
then there is a constant $c$ such that $f(x)=c$ for almost every $x\in (a,b)$.
I started by noting that $C_c^\infty([a,b])$ is dense in $L^1([a,b])$, so we can find a sequence $\{f_n\}\subseteq C_c[a,b]$ such that $\|f_n-f\|_{L^1[a,b]}\to 0$ as $n\to \infty$. If we have $(*)$ for $f_n$, where $n$ is sufficiently large, then we can conclude that $f_n$ is constant, and therefore $f$ is constant a.e.. But I got stuck when trying to show $(*)$ for $f_n$, $n$ large. It seems that the usual triangle inequality type argument wouldn't work here since there is a $\frac 1h$ in the front.
So I was wondering if there is any way to make the above argument work? Or if not, how to solve the problem? Any suggestions/hints would be appreciated!
| This is an easy consequence of Lebesgue's Theorem. Let $a<c<d<b$ and assume that $c$ and $d$ are Lebesgue points of $f$. Then $\frac {\int_c^{d} f(x+h)dx -\int_c^{d}f(x)dx} h \to 0$. This gives $\frac {\int_d^{d+h} f(x+h)dx -\int_c^{c+h}f(x)dx} h \to 0$. Since $c$ and $d$ are Lebesgue points this gives $f(c)=f(d)$. Since almost all points are Lebesgue points it follows that $f$ is a.e. constant
| {
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Prove the $\ell^2$ norm of a linear transformation $A: \mathbb{R}^n \to \mathbb R^n$ is the maximum eigenvalue If $A: \mathbb R^n \to \mathbb R^n$ is a linear transformation and $\mathbb R^n$ is equipped with $\lVert \cdot \rVert_2$, prove that
$$ \lVert A \rVert := \sup \left\{ \frac{\lVert A \vec{x} \rVert_2}{\lVert \vec{x} \rVert_2} : \vec{x} \in \mathbb R^n, \vec{x} \neq 0\right\} = \max \{ \lvert \lambda_i \rvert : i = 1, \ldots, n \} $$
where $\lambda_i$ is the eigenvalues of a matrix $B$ such that $A(\vec{x}) = B \vec{x}$ for every $\vec{x} \in \mathbb R^n$.
Above is the problem I'm struggling with. I don't know how to show this is true, I searched a lot about it, but didn't understand the concepts! Please help me understand and prove this!
| The transformation $T(1,0)=(1,0), T(0,1)=(1,1)$ is non-singular has only 1 as its eigen value but $||T||=\sqrt 2$. The stated equality holds for symmetric matrices (even singular ones!) but the result claimed is false.
| {
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Probability Random selection from class
*Suppose that a certain college class contains $62$ students. Of these, $35$ are sophomores, $38$ are biology majors, and $12$ are neither. A student is selected at random from the class.
(a) What is the probability that the student is both a sophomore and a biology major?
(b) Given that the student selected is a biology major, what is the probability that he is also a sophomore?
My answer:
a.$P(S and B)= (35/62)+(38/62)-((62-12)/62))=23/62$
b.$P(S|B)=(23/62)/(38/62)=(23/38)$
| I agree with @MattiP that your answers are correct. But I want to mention
a general method for problems like this. (Making a table of the kind I
suggest is also useful for somewhat more advanced topics in statistics.)
It is often useful to make a $2 \times 2$ table for such situations:
You are given this information:
SOPH
-------------------
BIOL Yes No Total
------------------------------------------
Yes 38
No 12
------------------------------------------
Total 35 62
From there you can find the marginal totals for both Biology and Sophomore.
Then because you have one count (12) in the body of the table, you can get
the rest.
Sometimes people say such a table has only 1 'degree of freedom' because
once you have the marginal information and one count in the body of the
table, the other three counts are determined.
When you are finding the conditional probability $P(S|B)$ you are only interested in the first row
of the table (for BIOL = Yes). Twenty-three of the 38 Biology majors are
Sophomores.
| {
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Decaying rate of a convolution between an integrable function and a Schwartz function Suppose $f\in L^1(R^n)$ and $g\in S(R^n)$, where $S(R^n)$ is Schwartz space. Then, Can I have estimation like following?
$$
|[f*g](x)|\leq\frac{1}{(1+|x|)^{s}},
$$
for some $s>n$.
If it is correct, how to prove it? If it is not correct, what additional assumptions does it require?
| Of course the answer to the question as stated is "of course not"; the sensible version of the question is whether we have $$
|[f*g](x)|\leq\frac{c}{(1+|x|)^{s}}.
$$
The answer to that question is still no, although it's not so obvious.
Take $n=1$ just to simplify the notation. Choose $g\in\mathcal S(\Bbb R)$ with $g\ge0$ and $g(x)\ge1$ for all $x\in[-1,1]$. Given a sequence $a_n\to\infty$, let $$f=\sum_{n=1}^\infty\frac1{n^2}\chi_{[a_n,a_n+1]}.$$
Then $f*g(a_n)\ge\frac1{n^2}$, so if $a_n$ blows up fast enough, in particular if $(1+|a_n|)^s/n^2\to\infty$, then $(1+|x|)^sf*g(x)$ is not bounded.
Note The same argument shows that $f*g\in C_0$ is the most that can be said about how fast $f*g$ vanishes at infinity: If $\phi\in C_0$ it is not true that $f*g=O(\phi)$ for every $f\in L^1$ and $g\in\mathcal S$.
| {
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The inexplicable approach of an Indian mathematician for Cosecant I am reading about the old Indian mathematician approximation
$$\csc (z)\simeq \frac{z^4+\pi ^2 z^2+2 \pi ^4}{2 \pi ^4 z-2 \pi ^2 z^3}$$
reminiscent of Bhaskara's.
I tried to use Taylor series and I got something similar to a Padé approximation
$$\csc (z)\simeq \frac{\frac{11 z^4}{5880}+\frac{3 z^2}{49}+1}{z-\frac{31 z^3}{294}}$$ but the Indian formula is a fairly superior approximation.
How is this possible?
| The Indian approximation encodes a large portion of the Weierstrass product for the sine function:
$$ \sin(z) = z\left(1-\frac{z^2}{\pi^2}\right)\prod_{n\geq 2}\left(1-\frac{z^2}{n^2\pi^2}\right) \tag{1}$$
The problem of approximating $\csc(z)$ boils down to producing approximations for
$$ g(z)=\frac{z}{\sin z}\left(1-\frac{z^2}{\pi^2}\right)=\prod_{n\geq 2}\left(1-\frac{z^2}{n^2 \pi^2}\right)^{-1}\tag{2}$$
which is an even and approximately quadratic function, ranging from $1$ to $2$ on the interval $(0,\pi)$.
Actually
$$ g(z) \approx 1+\left(\zeta(2)-1\right)\left(\frac{z}{\pi}\right)^2+(2-\zeta(2))\left(\frac{z}{\pi}\right)^4\tag{3} $$
leads to an approximation similar to the Indian one
$$ \csc(z)\approx\frac{1+\left(\zeta(2)-1\right)\left(\frac{z}{\pi}\right)^2+(2-\zeta(2))\left(\frac{z}{\pi}\right)^4}{z\left(1-\frac{z^2}{\pi^2}\right)}.\tag{4}$$
and more accurate if $z$ is close to $0$.
$$ \csc(z)\approx\frac{1+\left(\zeta(2)-1\right)\left(\frac{z}{\pi}\right)^2+(\frac{7}{2}-2\zeta(2))\left(\frac{z}{\pi}\right)^4+(\zeta(2)-\frac{3}{2})\left(\frac{z}{\pi}\right)^6}{z\left(1-\frac{z^2}{\pi^2}\right)}\tag{5}$$
is more accurate on the whole interval $(0,\pi)$. Folklore: the evaluation of $(5)$ at $z=\frac{\pi}{6}$ leads to
$$ \pi\approx \frac{3}{35}\left(432-\sqrt{156301}\right) $$
with an approximation error of $\approx 3$ parts in one hundred thousands.
| {
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Prove $\|A\| \leq \|A\|_{HS}$, where $\|A\|$ is the operator norm of A I Am trying to solve the following problem Let
$H_1
,H_
2$
be Hilbert spaces. Let
$A
\in B(H_1
,H_2
)$ be a Hilbert-Schmidt operator. For a complete
orthonormal sequence $(
u_n
)$ in
$H_1$, define the
Hilbert-Schmidt norm
$\|.\|_{HS}$
by
$\|A\|_{HS}=\left(\sum^\infty_{n=1}\|A(u_n)\|^2\right)^\frac{1}{2} $
Prove $\|A\| \leq \|A\|_{HS}$, where $\|A\|$ is the operator norm of A.
My current attempt is as follows:
Let $x \in H_1$ such that $ x=\sum^\infty_{n=1} \langle x,u_n\rangle u_N$ and $\|x\|=\left(\sum^\infty_{n=1}|\langle x,u_n\rangle|^2\right)^\frac{1}{2} $. Then we have,
$\|A(x)\|=\|A\left(\sum^\infty_{n=1} \langle x,u_n\rangle u_N\right)\| =\|\sum^\infty_{n=1}A\left( \langle x,u_n\rangle u_n\right)\| = \|\sum^\infty_{n=1} \langle x,u_n\rangle A\left(u_n\right)\| \leq \|\sum^\infty_{n=1}A(u_n)\|\ \|\sum^\infty_{n=1}\langle x,u_n\rangle\|$
I am unsure where to go from here and feel I have made a mistake? Thank you in advance.
| In the last inequalities you have used an inequality of the form:
$$\left\Vert\sum_n a _n b_n \right\Vert \leq \left\Vert\sum_n a _n \right\Vert \left|\sum_n b_n \right|$$
which is not true. However you can use a Cauchy-Schwartz inequality which gives:
$$\sum_n \left\Vert a _n\right\Vert |b_n| \leq \left(\sum_n \left\Vert a _n \right\Vert^2 \right)^\frac{1}{2} \left(\sum_n \left| b _n \right|^2 \right)^\frac{1}{2} $$
| {
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What is the contrapositive of this simple proposition?
If $a$ and $b$ are rational then $ab$ and $a+b$ are also rational.
What is the contrapositive of this proposition?
*
*If $ab$ and $a+b$ are not rational then $a$ and $b$ are not rational.
*If $ab$ or $a+b$ are not rational then $a$ or $b$ are not rational.
| Your statement: $a$ and $b$ rational $\implies$ ($ab$ and $a+b$) rational.
Contrapositive: not ($a+b$ and $ab$ rational) $\equiv$ not ($a+b$ rational) or not ($ab$ rational) $\implies$ not($a$ and $b$ rational) $\equiv$ not ($a$ rational) or not ($b$ rational).
(My initial answer had a silly mistake in it)
| {
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Show that if $Y$ is path connected, then there is only one homotopy class of maps of $[0,1]$ into Y This question is taken from Mukres' Topology (exercise 51.2.b). This site provides the following answer:
All “closed” segments of a path are homotopic to the whole path. Since $Y$ is path connected, for any two paths we can connect the final point of the first one with the initial point of the second one by a path to form the product of three paths, in which the two given paths are “closed” segments. Given the idea, it can be shown formally, of course.
Why does this prove that all paths in $Y$ are homotopic? For any two paths to be pasted to a new path in the same space does not prove that any two paths are homotopic, right?
My attempt at this problem is the following:
Take any two paths $f, f':[0,1] \rightarrow Y$, then $F(t,x) := (1-t)f(x) + tf'(x)$ is continuous and has $F(0,x) = f(x)$ and $F(1, x)=f'(x)$, thus $F$ is a homotopy between any two paths in $Y$ and thus all paths in $Y$ are homotopic.
However, I think that I am wrong since this doesn't use path-connectedness of $Y$.
| Write $I=[0,1]$. Then there is a homotopy between the constant
map $I\to I$ and the map $I\to I$ taking the whole of $I$ to $0$. Composing this with any
continuous map $f:I\to Y$ gives a homotopy between $f$ and the constant
map taking $I$ to $f(0)$. So every map from $I$ to $Y$ is homotopic
to a constant map. If $Y$ is path-connected, then any two constant
maps $I\to Y$ are homotopic.
| {
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Prove the following sequence diverges: $\sqrt{n}-\frac{1}{n^2}+4$ I am trying to work through proving some sequences diverge. I am having a really hard time with the inequality arguments and I'm not sure why. The current problem is proving that $$\sqrt{n}-\frac{1}{n^2}+4$$ diverges to infinity.
I understand that essentially I let $c$ be an arbitrary positive number and then I have to find some natural number $N$ dependent on $c$ so that $$\sqrt{n}-\frac{1}{n^2}+4>c$$ for all $n\geq N$. The trouble I have is sussing out what $N$ needs to be for an arbitrary chosen $c$.
| I think that I got it thanks to dem0nakos comment about not needing the best possible $N$.
Proof:
Let $c$ be any positive number. By the Archimedean property we can select a natural number $N_1$ so that $N_1>4c^2$ and therefore $\sqrt{N_1}>2c$. Simultaneously we can find an $N_2$ such that $\frac{1}{N_2^2}<c$. If we let $N=\max\{N_1,N_2\}$ and take $n\geq N$ then we have
$$\sqrt{n}-\frac{1}{n^2}+4>\sqrt{4c^2}-\frac{1}{n^2}+4=2c-c+4=c+4>c$$ as desired.
| {
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Tricky Graph Theory Puzzle I ran into an intriguing puzzle on Reddit that I thought could use some attention.
You start with a 3x3 grid labeled with numbers 1-9 like this:
$$ 1 \text{ }\text{ }\text{ }2\text{ }\text{ }\text{ }4
\\ 5 \text{ }\text{ }\text{ }6\text{ }\text{ }\text{ }8
\\ 9 \text{ }\text{ }\text{ }3\text{ }\text{ }\text{ }7$$
To complete the puzzle, you must connect directed edges between the numbers so that:
*
*The numbers connect in order from 1-9. There can be intermittent numbers when connecting them. For example, some puzzle may have the solution: 1, 3, 2, 9, 4, 3, 9, 4, 8, 5, 7, 8, 6, 7, 8, 2, 9.
*The edges cannot overlap each other.
*The vertices (the numbers) can have multiple edges connecting to it.
*You cannot go back over edges (no overlaps).
Here's a picture of a completed puzzle:
I have a couple of question relating to this problem:
If there are impossible puzzles, then what criteria determines if a puzzle is unsolvable? If I were to make it 4x4 or 5x5 or NxN, would the criteria extend to those? What proportion of puzzles for an NxN puzzle are solvable.
Here's the link to the Reddit post: https://www.reddit.com/r/math/comments/8b032u/need_help_for_an_eulerian_path_game_i_made/
| Condition 4: Not being allowed to go back over edges definitely changes the complexion of the problem. In fact, for ALMOST ALL numberings of an $n \times n$ grid as $n$ gets large, such a completion of the puzzle is impossible.
Why? The resulting digraph on the $n \times n = n^2$ vertices (vertices here taken as points on the grid) would be planar due to Condition 2. This would imply for some subset $S$ of $\frac{n^2}{2}$ vertices of the digraph, then for some $R$ no more than $4n$, there are exactly $R$ edges leaving $S$. (Theorem of Lipton-Tarjan I believe). However, [one can check that] for $n$ sufficiently large, a random bijective numbering $\cal{N}$ of the $n \times n$ vertices of the grid (where the numbers used are $1,2,3,\ldots, n^2\}$), would be such that there are many more than only $4n$ integers $i$ that satisfy the following: $i$ is in $S$ but $i+1$ is outside of $S$. However, one can check that if $\cal{N}$ is a numbering such that there are $R'> R$ integers $i$ such that $i$ is in $S$ and $i+1$ is not, then solving the puzzle with the numbering as $\cal{N}$ is impossible. Indeed, such a path would have to leave $S$ at least $R' > R$ times which means that an edge leaving $S$ would be repeated.
| {
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How can I define closeness of these geometric shapes Given points on a 2D plane, what kind of metrics can be used to define if they closely fit either:
*
*triangle
*square or rectangle
*circle
*oval (circular but not oval)
(Image credit: StyleCraze.com "How to Determine the Shape of Your Face".)
Note thanks to John Gowers on Meta for trying to help clarify parts of my
old question, which was closed.
| The four shapes given are all examples of (boundaries of) 2-dimensional convex bodies, so any metric on arbitrary convex bodies will do. Some examples include the following, where $C,D$ are $d$-dimensional convex sets in Euclidean space:
*
*Hausdorff metric: $d(C,D)= \max\left\{\sup\limits_{x \in X}\inf\limits_{y \in Y}~d(x,y),~\sup\limits_{y \in Y}\inf\limits_{x \in X}d(x,y)\right\}$ where $d(x,y)$ is the Euclidean distance function.
*Symmetric difference metric: $\Delta_v(C,D)= v (C \cup D) - v (C \cap D)$, where $v$ is Euclidean volume in $\mathbb{E}^d$. See page 1 of this paper and this section of the symmetric difference wiki page.
*Symmetric surface area deviation: $\Delta_s(C,D)= s(C \cup D) - s(C \cap D)$, where $s$ is the surface area. See page 1 of this paper.
Note: The symmetric surface area deviation does not satisfy the triangle inequality, so is not technically a metric, but rather a deviation measure.
Note: One can replace the continuous volume in (2) with discrete approximations and still have a metric.
| {
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Show that $p(x) = a_nx^n+a...+a_1x+a_0$ either has a root or attains minimum value in R. Show that if $p(x) = a_nx^n +\dots+ a_1x + a_0$ and $a_n > 0$, then either $p(x) = 0$ has a solution, or else $p(x)$ has attains minimum value on $\mathbb{R}$.
I'm sorry I don't know how to even start the problem. I know that if it is an odd degree polynomial then it has roots.
Thank you. The second part confuses me the most. This is from an undergrad Real Analysis homework.
| Hint: Recall that polynomials are continuous on $\Bbb R$; note that $p(x)\to\infty$ as $x\to\infty$; if $p(x)\lt 0$ for some $x=a$, apply IVT to $p$ on $[a,\infty)$; if $p(x)\gt 0$ for all $x$, recall that $\Bbb R$ has the glb property.
| {
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Question on irreducibility of $x^{nm}-a$ when $n$ and $m$ are coprime Let $F$ be a field and $a\in F$ and let $m,n$ be coprime positive integers such that. Then $x^{mn}−a$ is irreducible in $F[x]$ if and only if both $x^m−a $ and $x^n−a$ are irreducible in $F[x]$. This is supposed to be a Galois Theory question however I don't see how I can use Galois theory to prove such a result. I mean all I can say is that is that the splitting field of $x^{mn}-a$ is $F(z_n,c)$ where $z$ is a primitive nth root of unity and $c^n=a$ I have no idea how to continue from there any help/solution will be greatly appreciated.
Thanks in advance
| This is an exercise on irreducibility, not on splitting fields. If $X^{mn}-a$ is irreducible over $F$, so are $X^{m}-a$ and $X^{n}-a$ because $X^{mn}-a=(X^{m})^n-a=(X^{n})^m-a$.
Conversely, assume $X^{m}-a$ and $X^{n}-a$ irreducible over $F$, and let $\alpha$ be a root of $X^{mn}-a$ in an algebraic closure. Then $\alpha^m$ is a root of $X^{n}-a$, hence $[F(\alpha ^m) :F]= n$, and similarly $[F(\alpha ^n) :F]= m$. If $m$ and $n$ are coprime, a classical result on the multiplicativity of degrees in towers ensures that $[F(\alpha ^m, \alpha ^n):F]= mn$. But by Bezout's theorem, there exist integers $r, s$ s.t. $rm+sn=1$, so that $\alpha = \alpha^{rm} \alpha^{sn}$, and $F(\alpha ^m, \alpha ^n)=F$. Hence the degree of $\alpha$ over $F$ is $mn$, and $X^{mn}-a$ must be irreducible over $F$.
| {
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Showing the kernel of a set $X\subset \Bbb R^2$ is convex
Let $X\subset \Bbb R^2$. We define $V(x)$ to be the set of points in $X$ that $x$ 'can see', i.e. $V(x)=\{y\in X\mid [x,y]\subset X\}$.
The kernel is then $\text{ker}(X)=\{x\in X\mid V(x)=X\}$.
I want to show that the kernel is a convex set.
This is an exercise in a section of my textbook that covers only Helly's and Radon's theorems, so I would imagine they are relevant, but cannot work out how to use them. So I tried to just attack it directly:
If $\text{ker}(X)=\emptyset$ (the space is not star shaped), then this is convex vacuuously. If $|\text{ker}(X)|=1$ then this holds trivially. So say $\text{ker}(X)$ has at least two points $x,y$. We need to show that all the points $tx+(1-t)y$ are in $\text{ker}(X)$ for $t\in [0,1]$. For any other point $z\in X$, we have $[x,z]$ and $[y,z]$ contained in $X$. Then either $x,y,z$ are affine dependent, or $X$ contains the convex hull of $\{x,y,z\}$ in which case $\text{ker}(\text{convex}(\{x,y,z\}))=\text{convex}(\{x,y,z\})$ in which case $tx+(1-t)y$ can see $z$, for each $t\in[0,1]$. For any pair $x,y\in\text{ker}(X)$, we can do this for all $z\in X$, and so for each $t\in[0,1]$ we have that $tx+(1-t)y\in \text{ker}(X)$ so we are done.
I am not really sure if my argument is sound honestly, but maybe the approach is almost there?
Please tell me if my proof is OK, or otherwise give a nicer proof.
| If $a,\ b\in {\rm Ker}\ X$ and $z\in X$, then $$ [az],\ [bz]\subset
X $$
If $c=ta+(1-t)b,\ 0<t<1$, then assume that $[cz]$ is not in $X$.
That is, there is $x\in (cz)$ not in $X$.
Since $[az]\subset X$, so $[sa+(1-s)z\ b]\subset X$ by assumption on
$b$.
For suitable $s$, $[sa+(1-s)z\ b]$ contains $x$.
| {
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What's that process called where you form new group law by $x \star y = x \cdot a \cdot y$ for some $a$ in the group? Let $A$ be an abelian group. We can form new groups $(A, \cdot a \cdot)$ where $a$ is any element of $A$. Choosing $a = 1$ the identity of $A$ gives $A$ itself.
Clearly, associativity comes from associativity and commutativity of $A$.
Identity is $x \cdot a \cdot e = x$ or $e = a^{-1}$.
Inverse is $x \cdot a \cdot y = a^{-1}$ or $y = a^{-1} x^{-1} a^{-1}$.
What is this group formation process called and do the groups relate back to $A$?
I've seen it somewhere, and can't find where again.
| I have seen this also for groups, but it seems to be more interesting for Lie algebras, see here and here. For groups it seems to be called variant, see the comments at the above questions, or a sandwich, e.g., Semigroups under a sandwich operation, Proc. Edinburgh Math. Soc. (Ser. 2) 26 (1983), 371-382.
For $K$-algebras $(A,\cdot)$ with multiplication $x\cdot y$ a new multiplication
$x\circ_z y$ depending on a fixed $z$ is called a homotope, or a mutation.
| {
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For compass and straightedge problems, are you allowed to use the compass as a ruler? For compass and straightedge problems, you could have a line between two points A and B, and want to make a line the same size between C and line DE.
If you placed the two points of the compass between A and B, and made a circle around C with the same radius, that would achieve this result.
But is this something you are allowed to do?
| Yes. Not by the rules about how to use compass and straightedge but because it can be proved that it's as if we could do it (that's proposition 2 of book I of Euclid's Elements).
| {
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$C^*$-subalgebra of $M_2(\mathbb{C})$ Consider the following subalgebra of $M_2(\mathbb{C})$:
$$A=
\left\{\begin{bmatrix}
a & b \\
b & a
\end{bmatrix} : a,b\in \mathbb{C}\right\}.$$
One method I know that $A$ is isometrically *-isomorphic to $C(K)$, where $K=\{1,2\}$. (Because we can assume function $f$ such that $f(1)= \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}$
and
$f(2)=\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}$, then we use the Gelfand transform between $C(K)$ and $C(\mathscr{M})$, where $\mathscr{M}$ is the maximal ideals of $C(K)$. Hence, using the Gelfand transform $\widehat{f}(J)=f(x), J\in \mathscr{M}, x\in K $ to get the maximal ideals of $A$.
But this problem is asked to show that $\Delta=\{\phi_{1}, \phi_{2}\}$, where $\phi_{1}(y)=a+b, \phi_{2}(y)=a-b$. Here, $\Delta$ is the set of all complex homomorphism $A \to \mathbb{C}$.
$\textbf{How to solve this problem in this way? }$
| Suppose that $\pi:A\to\mathbb C$ is a homorphism.
If $\pi(I)=0$, then $\pi=0$. Otherwise, from $\pi(I)=\pi(I^2)=\pi(I)^2$, we deduce that $\pi(I)=1$. In any case, $\pi(aI)=a$. Also, since
$$
x=\begin{bmatrix} 0&1\\1&0\end{bmatrix}
$$
satisfies $x^2=I$, we deduce that $\pi(x)^2=1$. So $\pi(x)=1$ or $\pi(x)=-1$. And those are all the choice we have: when $\pi(x)=1$, we obtain the homomorphism
$$
\pi\left(\begin{bmatrix} a&b\\b&a\end{bmatrix} \right)=\pi(aI+bx)=a+b.
$$
When $\pi(x)=-1$, we obtain
$$
\pi\left(\begin{bmatrix} a&b\\b&a\end{bmatrix} \right)=\pi(aI+bx)=a-b.
$$
| {
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Problem on normed space. Let $X$ be a normed space, $Y$ a dense subspace of $X$ and $Z$ a closed finite-codimensional subspace of $X$. Is $Z\cap Y $dense in $Z$ ?
I have no idea how to solve this problem. I am using this website for the first time, any help would be appreciated.
| Suppose first that the codimension of $Z$ is $1$. Then $Z=\ker\phi$ for some nonzero $\phi\in X^*$. There is some $y\in Y$ such that $\phi(y)\neq0$. Define $P:X\to X$ by
$$Px=x-\frac{\phi(x)}{\phi(y)}y.$$
Then $P$ is linear and bounded, $PX=Z$, and $PY\subset Y\cap Z$. If now $z\in Z$, there is a sequence $\{y_n\}$ in $Y$ converging to $x$. Thus $\{Py_n\}$ is a sequence in $Y\cap Z$ converging to $Pz=z$.
In general, if $Z$ has codimension $n$, then $Z=\bigcap_k\ker\phi_k$ for some nonzero $\phi_1,\ldots,\phi_n\in X^*$, and we can find $y_1,\ldots, y_n\in Y$ such that $\phi_k(y_k)\neq0$. Then define $P:X\to X$ by
$$Px=x-\sum_{k=1}^n\frac{\phi_k(x)}{\phi_k(y_k)}y_k.$$
| {
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The case for left and right adjoint functors commuting with colimits and limits respectively when the functors are presheaves. Presheaves $F:\mathcal{C}^{opp} \rightarrow \mathcal{D}$, $G:\mathcal{D}^{opp} \rightarrow \mathcal{C}$ are called left (resp. right) adjoint, if there exist a natural bijection (in $C \in \text{Ob}\ \mathcal{C}$, $D \in \text{Ob}\ \mathcal{D}$)$$\text{Hom}_{\mathcal{C}}(G(D),C) \simeq \text{Hom}_{\mathcal{D}}(F(C),D) \ \ (\text{resp. }\ \text{Hom}_{\mathcal{C}}(C,G(D)) \simeq \text{Hom}_{\mathcal{D}}(D,F(C)))$$
Now, I know that left adjoint functor commutes with colimits and right adjoint functor commutes with limits. I wanted to know if anything changes if we consider functors of the sort above.
| A functor $G:\mathcal{D}^{op}\to\mathcal{C}$ can instead be considered as a functor $G^{op}:\mathcal{D}\to\mathcal{C}^{op}$, and $F$ and $G$ are left adjoint iff $F$ is left adjoint to $G^{op}$ in the usual sense. So this means $F$ preserves colimits and $G^{op}$ preserves limits, and the latter condition is equivalent to $G$ preserving colimits.
To be clear, when we say "$F$ preserves colimits", that means it turns colimits in $\mathcal{C}^{op}$ into colimits in $\mathcal{D}$. Thinking of $F$ instead as a contravariant functor from $\mathcal{C}$ to $\mathcal{D}$, this means it turns limits in $\mathcal{C}$ into colimits in $\mathcal{D}$. Similarly, $G$ will turn limits in $\mathcal{D}$ into colimits in $\mathcal{C}$.
Of course, the dual story holds when $F$ and $G$ are right adjoint: they both preserve limits, meaning that when you consider them as contravariant functors, they turn colimits into limits.
| {
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The longer the base, the longer the hypotenuse Excuse me if this is a silly question, but my plane geometry is very rusty. When I re-read Jack D'Aurizio's answer to the question "How can we prove that $\pi > 3$ using this definition", I wondered why, when viewed from the perspective of high school plane geometry, the area of the regular hexagon of unit spoke length is smaller than the area of the unit disc. Obviously, this follows if the regular hexagon entirely lies inside the unit circle, but why is this true? My first thought was that this is because the distance from the centre of the hexagon to any point on its perimeter is at most $1$. Put it another way, this is a consequence of the following (stronger) statement:
*
*Given $\bigtriangleup ABC$, if $X$ lies on the line segment $\overline{BC}$, then the length of $\overline{AX}$ is $\le$ the larger length of $\overline{AB}$ or $\overline{AC}$.
The above statement in turn follows from another statement:
*
*Let $\overline{AP}$ be a fixed line segment and $\bigtriangleup APX$ be a right-angled triangle, where $\overline{AP}\perp\overline{PX}$. The longer the base $\overline{PX}$, the longer the hypotenuse $\overline{AX}$.
This follows directly from Pythagoras theorem. However, all proofs of Pythagoras theorem that I knew, such as the ancient Chinese proof or Einstein's proof, make use of the notion of area. Why must the proof of a statement about length involve the concept of area?
So, here is my question:
*
*Can the statement in the second bullet point be proved, within the framework of Eucliedan plane geometry, without using Pythagoras theorem or the notion of area?
| Let extend $PX$ and take point $X_1$ such as $PX_1>PX$. Then it's easy to see that $\angle{AXX_1}$ is obtuse and $AX_1$ is the biggest side in $\triangle{AXX_1}$
| {
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Sub sub sequences and a relation between convergence in probability and a.s convergence I am trying to understand an answer to this question (specifically Siméon's answer) Convergence in probability of the product of two random variables
Im struggling with the statement "$X_{n}$ tends to X in probability if and only if , every subsequence of $X_{n}$ has a sub sub sequence that tends to $X$ a.s"
I felt like the above statement deserved a question in itself.
If this is true I can see how the proof follows.
| I'll list off three main results that are useful to prove this result. If you've seen these before then the result you which to prove is pretty simple. However if you haven't, I've included proofs of the results.
Claim 1: "$X_n$ tends to $X$ in probability if every subsequence of $X_n$ has a sub-subsequence that tends to $X$ in probability"
To prove this suppose not, then there exists $\epsilon > 0$ such that $\Pr(\lvert X_n - X \rvert \geq \epsilon) \not \to 0$. Therefore there exists $\delta > 0$ such that for all $N$ there exists $n \geq N$ such that $\Pr(\lvert X_n - X \rvert \geq \epsilon) \geq \delta$. This allows us to construct a subsequence $X_{n_k}$ such that $\Pr(\lvert X_{n_k} - X \rvert \geq \epsilon) \geq \delta$ for all $k$. But then no possible sub-subsequence of $X_{n_k}$ can tend to $X$ is probability. Contradiction.
Claim 2: "If $X_n$ tends to $X$ in probability then there is a subsequence of $X_n$ which tends to $X$ almost surely"
By convergence in probability it is possible to pick a subsequence $X_{n_k}$ such that for all $k$
$$ \Pr\left(\lvert X_{n_k} - X \rvert \geq \frac{1}{k}\right) \leq \frac{1}{2^k}$$
In particular $\sum_{k = 1}^{\infty} \Pr( \lvert X_{n_k} - X \rvert \geq \frac{1}{k}) \leq 1 < \infty$ so by the first Borel-Cantelli lemma we have that
$$ \Pr\left( \lvert X_{n_k} - X\rvert \geq \frac{1}{k}\ \text{for infinitely many $k$} \right) = 0 \implies \Pr \left( \lvert X_{n_k} - X\rvert < \frac{1}{k}\ \text{eventually} \right) = 1$$
Therefore almost surely there exists $K$ such that for all $k \geq K$ we have that $\lvert X_{n_k} - X \rvert < \frac{1}{k}$. This implies $X = \lim_{k \to \infty} X_{n_k}$ on this event. Therefore $X_{n_k}$ tends to $X$ almost surely.
Claim 3: "If $X_n$ tends to $X$ almost surely then $X_n$ tends to $X$ in probability"
Fix $\epsilon > 0$ and consider $\Pr(\lvert X_n - X \rvert \leq \epsilon)$. Then we use the result that for any sequence of events $(A_n)$ we have
$$ \liminf \Pr(A_n) \geq \Pr(\liminf A_n)$$
(this is a version of Fatou's lemma). Applying this gives
\begin{align*}
\liminf \Pr(\lvert X_n - X \rvert \leq \epsilon)
&\geq \Pr(\lvert X_n - X \rvert \leq \epsilon\ \text{eventually}) \\
&\geq \Pr\left(\lim_{n \to \infty} X_n = X\right) = 1
\end{align*}
Therefore in fact $\lim \Pr(\lvert X_n - X \rvert \leq \epsilon) = 1$. So $\lim \Pr(\lvert X_n - X \rvert > \epsilon) = 0$ as required to convergence in probability.
Final Result: "$X_n$ tends to $X$ in probability if and only if every subsequence of $X_n$ has a sub-subsequence that tends to $X$ in almost surely"
For the forwards direction if $X_n$ tends to $X$ in probability then every subsequence of $X_n$ still tends to $X$ in probability so by Claim 2 there is a sub-subsequence which converges to $X_n$ almost surely.
Conversely if every subsequence of $X_n$ has a sub-subsequence converging to $X$ almost surely then said subsubsequence converges to $X$ in probability by Claim 3. So $X_n$ tends to $X$ in probability by Claim 1. This concludes the proof.
| {
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Given $1\leq aA question that was in my calculus 1 test, I remember the question from the mind, I'm pretty sure I didn't miss anything:
Given $1\leq a<b\leq2$, Prove that $$\frac{log(b)-log(a)}{b^2-a^2}<\frac{1}{2}$$
I would like to know the correct form to solve these kinds of questions.
What I did is that I showed the "Edge Cases". First is when $b=max[1,2]$ and $a\to max[1,2]$, so actually $b=2 , a \to 2$. We get:
$$\frac{log(2)-log(a)}{2^2-a^2}=L'Hoptial's \frac{0}{0}=\frac{-\frac{1}{a}}{-2a}=\frac{1}{2a^2}\to\frac{1}{8}<\frac{1}{2}$$
I did the same for $a=1 , b\to1$ where I got the term is $\to{\frac{1}{2}}^{-}$ which is $<\frac{1}{2}$. Also $a=1 , b=2$, And it works. Then I said that these cases cover all other cases when $1\leq a<b\leq2$. Is it a correct way to show that? What's a proper way to prove that? Thanks.
| "Calculus" is the key word. :)
By the mean value theorem: there is $c\in(a,b)$ such that
$$
\frac{\log(b)-\log(a)}{b-a}\frac{1}{b+a}=\frac{1}{c}\frac{1}{b+a}<\frac{1}{1}\frac{1}{1+1}=\frac{1}{2}.
$$
The inequality is because $c>a\geq 1$ and $b+a>a+a\geq 1+1$.
| {
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Stuck on a recursively defined converging sequence problem. I am given a simple quadratic equation
$$x^2-x-c=0, x>0, c>0$$
and then we define a sequence $\{x_n\}$ with $x_1>0$ fixed and then, if $n$ is an index for which $x_n$ has been defined, we define
$$x_{n+1}=\sqrt{c+x_n}$$.
With that I am asked to prove that $\{x_n\}$ converges monotonically to the solution of the polynomial.
I've done quite a bit of scratch work. Obviously we can solve the quadratic and the positive solution is $\frac{1+\sqrt{5}}{2}$. I have an inkling that the equation is decreasing and so I tried working with $x_{n+1}-x_{n+2}$ to show that the difference is positive but I didn't come up with anything useful.
I did realize that if I simply write out the limit we see that
$$\lim_{n\rightarrow\infty}[(\sqrt{c+x_{n+1}})^2)-\sqrt{c+x_{n+1}}-c]=\lim_{n\rightarrow\infty}[x_{n+1}-\sqrt{c+x_{n+1}}]$$
So if we want this final limit to go to zero then all I really need is $x_n$ to be monotonically decreasing since it is clearly bounded below by zero since $c$ and $x_1$ were taken to be positive.
| Hint
$$x_{n+1}^2-x_n=c$$
$$x_{n+2}^2-x_{n+1}=c$$
so,
$$x_{n+2}^2-x_{n+1}^2-(x_{n+1}-x_n)=0$$
$$(x_{n+2}-x_{n+1})(x_{n+2}+x_{n+1})=(x_{n+1}-x_n)$$
suppose that $x_{N+1}<x_{N}$ for some $N$. What can you conclude?
After that, you have to study the relation between $x_1$ and $x_2$, which will depends on $c$.
| {
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Uniform continuity of $f(x) = x$ on $\Bbb R$ I keep being told how trivial uniform continuity is for $f(x) = x$ on $\Bbb R$, but sometimes those simplest things are hardest for me to see. I haven't been able to find a proof for this anywhere, most likely BECAUSE it is just so simple... in lecture it was given to us as fact. I'd like to see how to get it.
In my head it follows easily from the definition I was given:
"A function $f: D \to \Bbb R$ is said to be uniformly continuous on $D \iff \forall \epsilon >0 , \exists \ \delta$ such that $|x-y| < \delta \implies |f(x) - f(y)| < \epsilon$"
But I know I can't use that in place of a proof so it's making me wonder if I truly do know what the definition is telling me. I thought I could simply USE $|x-y| < \delta \implies |f(x) - f(y)| < \epsilon$ since $f(x) = x$ but I'm second guessing? That doesn't seem right...
| Let $\epsilon>0$. What we wish to show that there exists a $\delta >0$ such that when we assume $|x-y| < \delta$, this implies $|f(x)-f(y)|< \epsilon$. But, note that
$$|f(x) - f(y)|=|x-y|.$$
So the "trivial" part is that we just choose $\delta = \epsilon$, because then we have
$$|x-y| < \delta=\epsilon \implies |x-y|<\epsilon \implies |f(x) - f(y)|<\epsilon$$
as needed!
| {
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Prove or disprove $ABA^{T} = AC$ implies $BA^{T} =C$ Suppose A is $m \times n$ non-zero matrix, $B$ is $n \times n$ of full rank and $C$ is $n \times m$ of full column rank. Can we prove or disprove that $ABA^{T} = AC$ implies that $C = BA^{T}$.
Note: Only $B$ is a square matrix. Rest are rectangular matrices with $m < n$.
| $A=\begin{pmatrix}1 & 0\\
0 & 0
\end{pmatrix}$, $B=C=I_{2}$
$ABA^{T}=\begin{pmatrix}1 & 0\\
0 & 0
\end{pmatrix}=AC$ but $BA^{T}=\begin{pmatrix}1 & 0\\
0 & 0
\end{pmatrix}\neq C$
| {
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Domains of higher powers of two unbounded self-adjoint operators Let $A : D(A) \rightarrow H$ and $B : D(B) \rightarrow H$ be two unbounded self-adjoint operators that are densely defined on a Hilbert space $H$. Suppose we know that $D(A) = D(B)$. Are there sufficient conditions for the equality $D(A^k) = D(B^k)$ to be true for positive integers all integers $k \in \mathbb{Z}_+$? Is there an example where this equality fails?
I guess this is true if $A$ and $B$ also commute with each other. But is there a more general sufficient condition?
| I realized there is an example why this could be generally hard.
Let $A :D(A) \rightarrow L^2(0,1)$ with $D(A) = \lbrace u \in H^2(0,1); u_x(0)=u_x(1)=0 \rbrace$ where $H^2(0,1)$ is the set of elements $L^2(0,1)$ that are least twice weakly differentiable. Additionally, for an element $a \in L^{\infty}(0,1)$ we define the multiplication operator $M_a : L^2(0,1) \rightarrow L^2(0,1)$ as $(M_af)(x)= a(x)f(x)$ for a.e $x \in (0,1)$. Then $D(A) = D(A-M_a) \subset H^2(0,1)$. Additionally, $D(A)^2 \subset H^4(0,1)$ but $D((A-M_a)^2) \not\subset H^4(0,1)$. For example, if $\mathbf{1}$ is the function taking value $1$ everywhere on $(0,1)$, then $\mathbf{1} \in D(A) = D(A-M_a)$, and $\mathbf{1} \in D(A^2)$. But $\mathbf{1} \notin D((A-M_a)^2)$ if $a$ is non-differentiable. The situation doesn't change if we replace $A$ by $A+\mathbb{I}$ with $\mathbb{I}$ being the identity operator.
| {
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Using complex exponential to show the indefinite integration of sin(x)sinh(x) dx Use the complex exponential to evaluate the indefinite integral of
$\sin x \sinh x$.
Express your answer in terms of trigonometric and/or hyperbolic functions
The attached photo is what I have tried so far
| Start with $\sin x=\Im e^{ix}.$ Then your integral is the imaginary part of the following integral:
$$\int e^{ix}\cdot\frac{e^x-e^{-x}}2\, dx=\frac 12 e^{ix}\cdot\left(\frac {e^x}{i+1}-\frac{e^{-x}}{i-1}\right)=\frac 12e^{ix}\cdot \frac{-(e^{-x}+e^x) + i(e^x-e^{-x})}{-2}=\\=\frac 12e^{ix}\cdot (\cosh x-i\sinh x).$$
As we are interested in the imaginary part, we can take use of the fact that $\Im ab=\Im a\cdot \Re b+\Re a\cdot \Im b$ and get the final result:
$$\frac 12\left(\sin x\cosh x-\cos x\sinh x\right)$$
It agrees with the result from WolframAlpha.
Disclaimer: I omitted $+C$ for clarity. Many people dislike it, so consider adding it.
| {
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Classifying 2-dimensional representations of $S_3$ by hand following Fulton & Harris In §1.3 of Fulton & Harris the authors guide the reader through a classification of the linear representations of $S_3$. Given a representation $\rho:S_3\to \mathrm{GL}(W)$ they first restrict to the abelian subgroup $A_3$ and use the fact it's a finite abelian group to conclude the image of its generator $\tau$ (rotation by a cube root of unity) decomposes $W$ into its eigenspaces. If I understand correctly, this is because $\tau$ is of finite order whence its minimal polynomial divides a separable polynomial and is thereby itself separable, which over $\mathbb C$ is equivalent to diagonalizability.
The authors moreover write the all eigenspaces are actually 1-dimensional.
Later on, to understand $\rho$ we are left with understanding how the second generator $\sigma$ (reflection) interacts with the eigenspaces of $\tau$. A calculation using the relations of $S_3$ (viewed as $D_3)$ proved that if $v$ is a $\tau$-eigenvector with eigenvalue $\lambda$, then $\sigma v$ is a $\tau$-eigenvector with eigenvalue $\lambda ^2$. From there we are left to check cases by possible $\tau$-eigenvalue $\lambda$.
In the case $\lambda=1$ there's a separation of cases I don't understand. It is claimed that given a $\tau$-eigenvector $v$ of eigenvalue $1$, the vectors $v,\sigma v$ may or not be independent. How can this be if the eigenspaces of $\tau $ are 1-dimensional (as suggested by the penultimate centered equation on p9)? The vectors $v,\sigma v$ are both $\tau$-eigenvectors of eigenvalue 1.
| The $\alpha_i$ on page 9 don't have to be distinct, so the eigenspaces don't have to be one-dimensional, a priori (and, in fact, since $W$ is an arbitrary representation there, they will not be one-dimensional in general).
| {
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Joint pdf of uniform dependent random variables Take $n$ non-negative dependent random variables $X_1,...,X_n$ with $Pr(X_i \leq t) = t, t\in[0,1]$ for every $i$ (uniform marginal distributions).
What is an example of a joint pdf for $X_1,...,X_n$ (with the given common marginal distribution), such that $E[\min_i X_i] = 1/2$?
| Assume that $(X_1,X_2,\ldots,X_n)$ is a solution and let $M=\min\{X_1,X_2,\ldots,X_n\}$.
Then, for every $i$, $M\leqslant X_i$ almost surely and $\mathbb E(M)=\frac12=\mathbb E(X_i)$ hence $M=X_i$ almost surely. This holds for every $i$ hence $X_1=X_2=\cdots=X_n$ almost surely.
In which case, naturally, $\mathbb E(M)=\frac12$ indeed holds, and $(X_1,X_2,\ldots,X_n)$ has no joint PDF.
On the other hand, for every positive $\epsilon$, there exists $(X_1^{(\epsilon)},X_2^{(\epsilon)},\ldots,X_n^{(\epsilon)})$ with uniform marginals and with a joint PDF, such that $\mathbb E(M^{(\epsilon)})\geqslant\frac12-\epsilon$.
| {
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Sum and Squares of Uniform Random Variables Suppose I have $X\sim U(-1,1)$ and $Z\sim U(0,0.1)$ independently and want to work out the pdf of a new r.v. defined as such: $Y=X^2+Z$. How could I calculate its distribution? I have tried to work it out but seem to have contradictory things appearing.
I have been able to show that the distribution of $Y|X=x \sim U(x^2,x^2+0.1)$ and hence we can deduce the joint pdf of $X$ and $Y$ is given by $$f_{XY}(x,y)=f_{Y|X}(y|x)f_X(x)=10\, \cdot \frac{1}{2}=5$$ and this is valid over the support $\{(x,y):-1<x<1,\, x^2<y<x^2+0.1\}$.
Hence to find $f_Y$ all that I surely need to do is integrate the joint pdf above over its support of $X$, ie. $\{x:-1<x<1\}$.
Doing this we get that $f_Y(y)=10$, which is valid for $\{y:0<y<1.1\}$, but how can this be? Where is the error in my reasoning?
| I prefer to handle cumulative distribution functions, rather like this:
For $0\le y \le 0.1$ you can say $$\mathbb P(Y \le y) = \int_{z=0}^y 10\, \mathbb P(X^2 \le y-z) \,dz = \int_{z=0}^y 10\, \sqrt{y-z} \,dz = \frac{20}{3}y^{3/2}$$
and taking the derivative will give you $f_Y(y)=10 \sqrt{y}$.
For $0.1\le y \le 1$ you can say $$\mathbb P(Y \le y) = \int_{z=0}^{0.1} 10\, \mathbb P(X^2 \le y-z) \,dz = \frac{20}{3}y^{3/2} - \frac{20}{3}(y-0.1)^{3/2}$$
and taking the derivative will give you $f_Y(y)=10 \sqrt{y}-10 \sqrt{y-0.1}$.
For $1\le y \le 1.1$ you can say $$\mathbb P(Y \le y) = \int_{z=0}^{y-1} 10 \,dz + \int_{z=y-1}^{0.1} 10\, \mathbb P(X^2 \le y-z) \,dz = 10y -\frac{10}{3} - \frac{20}{3}(y-0.1)^{3/2}$$
and taking the derivative will give you $f_Y(y)=10-10 \sqrt{y-0.1}$.
The cumulative distribution function turns out to look like
and the probability density function (zero outside $[0,1.1]$) looks like
| {
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Algebraic ideals inside $A \otimes \mathbb K$ Let $A$ be a unital C*-algebra and denote by $F(H)$ the algebraic ideal of finite rank operators on $H = \ell^2(\mathbb N)$. Is is true that
$$
A \odot F(H)
$$
is an algebraic ideal inside $A \otimes \mathbb K$ ? Here $A \odot F(H)$ denotes the algebraic tensor product of $A$ and $F(H)$.
I have the feeling that this is not true, since it might happen (?) that that the product $(a \otimes f)x$ with $a \in A, \ f \in F(H)$ and $x \in A \otimes \mathbb K$ is not an element of the algebraic tensor product but rather an element of the closure.
| I think your intuition is correct, even in the case of Abelian $C^\ast$-algebras. Take $A=c_0(\mathcal{Z})$. We have
$$
A \otimes K(H) \simeq c_0(\mathbb{N}; K(H)).
$$
For any element $x=(x_i)_i \in c_0(\mathbb{N}; K(H))$ we can define the following quantity
$$
\mathrm{Im}(x) = \overline{\mathrm{span}}\big\{ \bigcup_{i} \mathrm{Range}(x_i) \big\} \leq H.
$$
That quantity always gives a finite dimensional subspace if $x \in A \odot F(H)$. It only rests to see that there are cases in which $\mathrm{Im}(x \, f)$ is infinite, with $f \in A \odot F(H)$ and $x \in A \otimes K(H)$. Choose an ONB base $(e_i)_i$ of $H$ and denote by $(e_{i,j})$ the matrix units of $B(H)$. If $f_i = (i+1)^{-1} \otimes e_{1,1}$ and $x_i = (i+1)^{-1} \otimes e_{i,1}$, then $\dim \mathrm{Im}(x \, f) = \infty$.
Related: You can get a well-behaved algebraic ideal using the language of Hilbert $C^\ast$-modules. Take $A \odot H$ with the product given by $\langle a \otimes \xi, b \otimes \eta \rangle_X = a^\ast b \langle \xi, \eta \rangle$. And your ideal will be spanned by operators $\theta_{y,z}(x) = \langle y, x \rangle z$, see [Lance; p9].
Lance, E. Christopher, Hilbert $C^\ast$-modules. A toolkit for operator algebraists, London Mathematical Society Lecture Note Series. 210. Cambridge: Univ. Press,. ix, 130 p. (1995). ZBL0822.46080.
| {
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An affine bundle has a global section? Let $X$ be a manifold. We say $\pi: Y \longrightarrow X$ is an rank $n$ affine bundle if there is an open cover $\{ U_\alpha \}$ of $X$ such that
$ Y \big|_{U_\alpha} \cong U_\alpha \times \mathbb{R}^n $ and the transition function from $U_\alpha$ to $U_\beta$ is given by
$$ (x,v) \mapsto (x, \rho_{\beta \alpha }(x) v + u_{ \beta \alpha} (x)) $$ satisfying the cocycle condition
$ \rho_{\gamma \alpha} (x) = \rho_{\gamma \beta} (x) \rho_{\beta \alpha } (x) $ and
$u_{\gamma \alpha}(x) = \rho_{\gamma \beta} (x) u_{\beta \alpha} (x) + u_{\gamma \beta}(x)$.
Wikipedia claims that an affine bundle has a global section so it can be identified
with the vector bundle glued by the cocycles $\{ \rho_{\gamma \alpha} \}$ in a non-canonical way.
How can we construct one exactly? Someone claimed that local sections exist so one can glue them to a global one by
standard partition of unity argument. Since multiplying by constant doesn't make sense for affine bundle, I cannot see why this is obvious.
| Multiplication by functions doesn’t in general make sense, but affine combinations of sections do.
Specifically, if $\sigma,\tau\in\Gamma(\pi)$ are sections and $f, g\in\mathrm C^\infty(X)$ are smooth functions on the base summing to unity, then the obvious definition of $f\sigma + g\tau$ is independent of trivialization if (!) $f(x)+g(x) = 1$. This defines affine combinations only for a finite number of terms, but you can of course also do that for an arbitrary one, as long as only a finite number of coefficients are non-zero at any given point.
Now choose a (locally finite) trivializing open cover of the bundle, a partition of unity corresponding to that cover and an arbitrary local section over every neighbourhood. Then the affine combination of these sections with coefficients given by this partition is a well-defined global section.
| {
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minimum number $\alpha$ such that for all $x$: $\alpha^x \geqslant x$ This is a short question which I'm asking just out of curiosity:
Find the minimum positive number $\alpha \in \mathbb{R}$ such that $\forall x \in \mathbb{R}: \alpha^x \geqslant x$.
Maybe somebody even knows how it relates to other mathematical constants, e.g. to $e$.
You're also welcome with some hints or references to the corresponding literature :).
| This is only to summarize the answers:
The inequality holds for all $x \leqslant 0$ irrespectively to $\alpha$.
Assuming $x > 0$ we can apply the following transformations:
\begin{gather*}
\alpha^x > x \\
x\ln{\alpha} > \ln{x} \\
\ln{\alpha} > \frac{\ln{x}}{x} \\
\alpha > e^{\frac{\ln{x}}{x}}
\end{gather*}
If we want $\alpha$ to be greater than $e^{\frac{\ln{x}}{x}}$ for any $x$ we need to make it greater than maximum of $e^{\frac{\ln{x}}{x}}$.
This maximum is some number $M$ and we need to find minimum possible $\alpha$ such that $\alpha > M $.
This is impossible for real numbers $\mathbb{R}$ because for any number $\alpha$ we can find infinitely many numbers $\beta$ such that $M < \beta < \alpha$ - there is no minimum possible number greater than $M$. That is why J.G. was writing that my question was posed incorrectly and that I needed to replace $a^x > x$ with $a^x \geqslant x$.
Let's fix this
\begin{gather*}
\alpha^x \geqslant x \\
\ldots \\
\alpha \geqslant e^{\frac{\ln{x}}{x}}
\end{gather*}
Now we can try to find $M$:
\begin{gather*}
\left(e^{\frac{\ln{x}}{x}}\right)' = 0 \\
\left(\frac{\ln{x}}{x}\right)' = 0 \\
1 - \ln{x} = 0 \\
x = e;
\end{gather*}
which gives
\begin{equation*}
M = e^{\frac{\ln{e}}{e}} = e^{\frac{1}{e}}
\end{equation*}
The number $\alpha$ which we are seeking is the minimum possible number $\alpha \geqslant M$ - this is indeed $M$.
Therefore
\begin{equation*}
\alpha = e^{\frac{1}{e}}
\end{equation*}
| {
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Proving that if $\lim\limits_{n\to\infty}a_n=+\infty$ and $\{b_n\}$ is a bounded sequence, then $\lim\limits_{n\to\infty}(a_n+b_n)=+\infty$ I am trying to prove the following problem
If $\lim\limits_{n\to\infty}a_n=+\infty$ and $\{b_n\}$ is a bounded sequence, then $\lim\limits_{n\to\infty}(a_n+b_n)=+\infty$
I have these definitions as tools;
Definition 1. The sequence $\{a_n\}\to \infty$ if $\forall \;M\in R,\;\;\exists\;n_0=n_0(M)\in N$ $\ni$
$$n\geq n_{0}\;\implies\;a_n>M.$$
Definition 2. The sequence $\{b_n\}$ is said to be bounded if there exists $M>0,$ $\ni$
$$|b_n|\leq M \;\forall \;n\geq 1.$$
Can anyone help me out?
| Let $B$ be the bound for $(b_n)$. For any $M\gt0$ choose $n_0\in\mathbb N$ such that $n\ge n_0 \implies a_n\gt M+B$. Then $a_n+b_n\gt M$...
| {
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chain equivalence between cubical chain complex and simplicial chain complex I remember hearing before that for a topological space, the cubical chain complex and the simplicial chain complex are chain equivalent. Is this true? If yes, can someone provide me with a reference where I could see how this chain equivalence is constructed?
A natural choice /guess would be the chain map going from the cubical chain complex to the simplicial one (because cubes could be dissected into simplicies), however i have no clue about the reverse chain map.
Thank you
| This equivalence was first proved in the classical paper
Eilenberg, S., and Mac Lane, S., Acyclic models. Amer. J. Math. 75 (1953)
189–199.
The method of acyclic models was extended from chain complexes to crossed complexes in Section 10.4 of the book Nonabelian Algebraic Topology. The advantage of the more general case is that it takes into account the fundamental groups and groupoids and their actions and so more readily yields weak equivalences of spaces.
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"answer_id": 0
} |
Show there are infinitely many positive integers $k$ such that $\phi(n)=k$ has exactly two solutions where $n$ is a positive integer
Show that there are infinitely many positive integers $k$ such that the equation $\phi(n)=k$ has exactly two solutions, where $n$ is a positive integer.
Not entirely sure where to start or finish with this problem. I know that $\phi$ notation states that $\phi(n) = n(1-\frac{1}{p_1})(1-\frac{1}{p_2})\cdots(1-\frac{1}{p_S})$
but other than that I am lost.
| Let $k = 2 \times 3^{6m + 1}$. We claim that if $m > 0$, then $\phi(n) = k$ has exactly $2$ solutions. (For $m = 0$, it has four solutions: $7$, $9$, $14$, and $18$.)
Let $n = p_1^{a_1} p_2^{a_2} \cdots p_i^{a_i}$, where $p_j$ is a prime number, and $a_j$ is a positive natural number. Then $n$ is a solution if and only if
$$
p_1^{a_1 - 1} p_2^{a_2 - 1} \cdots p_i^{a_i - 1} (p_1 - 1)(p_2 - 1) \cdots (p_i - 1) = k.
$$
We note that if $n$ has at least two odd prime factors $p$ and $q$, then $(p - 1)(q - 1)$ is a factor of $\phi(n)$. Hence $4$ is a factor of $\phi(n)$, and sol $n$ is not a solution.
Thus any solution $n$ is of the form $n = 2^\alpha \cdot p^\beta$ where $p$ is some odd prime number, and $\alpha$ and $\beta$ are natural numbers, possibly equal to $0$. If $\alpha > 2$, then $\phi(8) = 4$ is a factor of $\phi(n)$, and so $n$ is not a solution. Thus $\alpha \in \{0, 1, 2\}$.
Suppose that $\alpha = 2$. Then if $\beta > 0$, we have that $2(p - 1)$ is a factor of $\phi(n)$, and so $4$ is a factor of $\phi(n)$, and so $n$ is not a solution. Thus if $\alpha = 2$, then $\beta = 0$, and so $n = 4$. But $\phi(4) = 2$ is not divisible by $3$, and so it is not equal to $2 \times 3^{6m + 1} = k$.
We thus must have that either $n = p^\beta$, or $n = 2 p^\beta$. In each case, we have that
$$
\phi(n) = p^\beta - p^{\beta - 1}.
$$
We thus must solve the equation
$$
p^{\beta - 1} (p - 1) = 2 \times 3^{6m + 1}.
$$
If $\beta = 1$, then this is equivalent to $p - 1 = 2 \times 3^{6m + 1}$. Not we note that $2 \times 3^{6m + 1} + 1 \equiv 2 \times 3 + 1 \equiv 0 \pmod 7$, and so this implies that $p = 7$. This corresponds to $m = 0$. We're considering the case where $m > 0$, so we can thus assume from now on that $\beta > 1$.
Since $\beta > 1$, we note that $p \mid 2 \times 3^{6m + 1}$, and so we have that $p = 3$. We thus must solve the equation
$$
2 \times 3^{\beta - 1} = 2 \times 3^{6m + 1}.
$$
This clearly has the unique solution $\beta = 6m + 2$, and so the two solutions to the equation $\phi(n) = k$ are given by $n = 3^{6m + 2}$, and $n = 2 \times 3^{6m + 2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2734926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
For a group $G$ such that $|G| = p^3$, for $p$ prime, either $|Z(G)| = p$ or $G$ is abelian. I came with this proof and I found some other proofs online, but mine is different and I want to see if I made any mistakes.
Problem:
Suppose $|G| = p^3$, where $p$ is a prime. Show that either $|Z(G)|=p$ or $G$ is abelian.
Case 1: $G$ is not abelian
We have the class equation
$$|G| = |Z(G)| + \sum_{g\in G\setminus Z(G) } \frac{|G|}{|\mathrm{Cent}(g)|}$$
where $Z(G)$ is the center of the group, and $|\mathrm{Cent} (g)|$ is the centralizer of $g$.
if we solve for $|Z(G)|$ we get
$$|Z(G)| = |G| - \sum_{g\in G\setminus Z(G) } \frac{|G|}{|\mathrm{Cent}(g)|} = p^3 \bigg( 1-\sum \frac{1}{\mathrm{Cent}(g)} \bigg)$$
Also, since $Z(G) \leq G$, by Lagrange's theorem, $|Z(G)|$ divides $|G| = p^3 = ppp$
Therefore, we have 4 possibilities: $|Z(G)| = \{1,p,p^2,p^3\}$
It can't be $p^3$ since that implies that $G$ is abelian, contradicting the initial assumption.
There's another theorem that states that if $|G|$ is a power of a prime number, then the center of $G$ contains nonidentity elements, so it can't be $1$.
This leaves us with $p$ or $p^2$.
If the size is $p^2$, we can go back to the class equation and obtain
$$p^3 = p^2 + \sum\frac{p^3}{|\mathrm{Cent}(g)|}$$
$$p\bigg( 1 - \sum\frac{p}{\mathrm{Cent}(g)}\bigg) = 1$$
$$\sum\frac{p}{\mathrm{Cent}(g)} = \frac{p-1}{p}$$
This number has to be an integer greater than or equal to $0$, and the only option is then $p=1$, but we already discarded that option, and $|Z(G)| \neq p^2$
Therefore, $|Z(G)| = p$ if $G$ is nonabelian.
If it is abelian then $G \setminus Z(G) = \emptyset$ so the sum is over no elements giving $0$, and $Z(G)=G \leq G$.
| Suppose that $G$ is non-abelian.
We know that $Z(G)\leq G$.
By Lagrange's Theorem $|Z(G)|$ must divide $|G|$.
Since $|G|=p^{3}$ the only possibilities are $1, p, p^{2}, p^{3}$.
$|Z(G)|\neq p^{3}$ because otherwise we will have $Z(G)=G$ but $G$ is non-abelian.
$|Z(G)|\neq p^{2}$ also because otherwise we will have the order of the factor group by the center as $|G/Z(G)|=|G|/|Z(G)|= p^{3}/p^{2} = p$. Therefore:
$|G/Z(G)|=p \implies G/Z(G)$ is cyclic $\implies G$ is abelian. But $G$ is non-abelian.
Now $|Z(G)|\neq 1$ also because $G$ is a $p-group$ and $p-groups$ have non-trivial center.
So, it must be that $|Z(G)|=p$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Minimum value trace of A*Atranspose For any matrix $A$, let $A^T$ denote its transpose matrix.
What is the minimum value of $\mathrm{trace}(AA^T)$ for an $n \times n$ non-singular matrix $A$ with integer entries?
| When $A$ is nonsingular, $AA^T$ is positive definite. Hence it has a positive diagonal. But $AA^T$ is also an integer matrix. Hence its diagonal entries are positive integers, meaning that $\operatorname{tr}(AA^T)\ge n$. Obviously, tie occurs when $A=I$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Is there an example of a, non-Hausdorff, topological vector space which has bounded subspaces Let $E$ be a topological vector space which is not Hausdorff.
Is it true that every non-trivial subspace $M$ of $E$ is necessarily unbounded?
Or, does there exist, non-Hausdorff, topological vector spaces containing non-trivial bounded subspaces? If any, what would be a counterexample?
Thanks in advance.
| Take $\mathbb{R}^2$ with the semi-norm $$\|(x,y) \| := |x|$$ and the subspace $$Y:= \{(0,y) : y \in \mathbb{R} \}, $$ then $( \mathbb{R}^2, \| \cdot \|)$ is a non-Hausdorff TVS with bounded subspace $Y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why can't I cancel $2x-3$ from $(2x-3)(x+5)=9(2x-3)$? Why are these simplifications wrong?
$$\begin{align}
(2x-3)(x+5)=9(2x-3) &\quad\to\quad \frac{(2x-3)(x+5)}{2x-3} = \frac{9(2x-3)}{2x-3} \quad\to\quad x+5 = 9\\[4pt]
x(x+2)=x(-x+3) &\quad\to\quad \frac{x(x+2)}{x} = \frac{x(-x+3)}{x} \quad\to\quad x+2=-x+3
\end{align}$$
| From here
$$(2x-3)(x+5)=9(2x-3)$$
we can observe that $2x-3=0$ is a solution and for $2x-3\neq 0$ we can cancel out and obtain
$$(2x-3)(x+5)=9(2x-3)\iff x+5=9\iff x=4$$
thus the solutions for the original equation are $x=\frac32$ and $x=4$.
As an alternative note that
$$(2x-3)(x+5)=9(2x-3)\iff 2x^2+7x-15=18x-27 \iff2x^2-11x+12=0$$
and then
$$x_{1,2}=\frac{11\pm\sqrt{121-96}}{4}=\frac{11\pm5}{4}=\frac32,4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
The Chain Rule - finding differentials I have been given:
*
*$V=f(x,y,z)$, with $x= r\cos\theta$, $y= r\sin\theta$ , and $z=t$.
*And asked, find $dV/dr$, $dV/d\theta$ and $dV/dt$
Would
$$
\frac{dV}{dr} = \frac{dV}{dx} * \frac{dx}{dr} + \frac{dV}{dy} * \frac{dy}{dr}
$$
?
If so, what is $dV/dx$ if I have just been given $V=f(x,y,z)$?
I've obtained:
$$V_{r}=V_{x}\cos(\theta )+V_{y}\sin(\theta )$$
I am unsure how to display my final answer.
| First: using the same name for different things is a bad habit that causes confusion. Use different names for different things. In your case:
$$V(r,\theta,t) = f(r\cos\theta,r\sin\theta,t).$$
Then:
$$
\frac{\partial V}{\partial r} =
\frac{\partial f}{\partial x}\frac{\partial x}{\partial r} +
\frac{\partial f}{\partial y}\frac{\partial y}{\partial r} +
\frac{\partial f}{\partial z}\frac{\partial z}{\partial r} = \cdots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Properties of a cumulative distribution function and rescaling Consider a cumulative distribution function $F$.
Take $a: (0,\infty) \rightarrow (0,\infty)$ and $b: (0,\infty) \rightarrow (-\infty,\infty)$. Let $a_s\equiv a(s)$ and $b_s\equiv b(s)$ for all $s \in (0,\infty)$.
Assume
$$
[F(a_sx+b_s)]^s=F(x), \quad\forall s>0
$$
could you help me to understand how this implies that
*
*$$ [F(a_{st}x+b_{st})]^s=F\left(\frac{a_{st}}{a_s}x+\frac{b_{st}-b_s}{a_s}\right), \quad \forall s,t>0 $$
*$$\left[F\left(\frac{a_{st}}{a_s}x+\frac{b_{st}-b_s}{a_s}\right)\right]^t=F\left(\frac{a_{st}}{a_ta_s}x+\frac{b_{st}-b_s-a_sb_t}{a_sa_t}\right). \quad\forall s,t>0 $$
I don't understand how things are rescaled.
| Because $$
(F(a_s x + b_s))^s = F(x), \quad \forall s > 0, x \in \mathbb{R}
$$
then for any $u > 0$, $y \in \mathbb{R}$, take $(s, x) = \left( u, \dfrac{y - b_u}{a_u} \right)$ to get$$
(F(y))^u = F\left( \frac{y - b_u}{a_u} \right).
$$
Now, for any $s, t > 0$, $x \in \mathbb{R}$, take $(u, y) = (s, a_{st} x + b_{st})$ to get$$
(F(a_{st} x + b_{st}))^s = F\left( \frac{a_{st} x + b_{st} - b_s}{a_s} \right) = F\left( \frac{a_{st}}{a_s} x + \frac{b_{st} - b_s}{a_s} \right),
$$
and take $(u, y) = \left( t, \dfrac{a_{st}}{a_s} x + \dfrac{b_{st} - b_s}{a_s} \right)$ to get\begin{align*}
\left( F\left( \frac{a_{st}}{a_s} x + \frac{b_{st} - b_s}{a_s} \right) \right)^t &= F\left( \frac{1}{a_t} \left( \left( \frac{a_{st}}{a_s} x + \frac{b_{st} - b_s}{a_s} \right) - b_t \right) \right)\\
&= F\left( \frac{a_{st}}{a_s a_t} x + \frac{b_{st} - b_s - a_s b_t}{a_s a_t} \right).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Axis Rotation to determine new points I am looking at this answer .
https://math.stackexchange.com/a/62248/474907, but the formula
c = cos(a); // compute trig. functions only once
s = sin(a);
xr = xt * c - yt * s;
yr = xt * s + yt * c;
differs from the wikipedia entry. So, I am a little confused, which one is correct.
$R_{z}(\theta )={\begin{bmatrix}\cos \theta &\sin \theta &0\\[3pt]-\sin \theta &\cos \theta &0\\[3pt]0&0&1\\\end{bmatrix}}$
https://en.wikipedia.org/wiki/Rotation_of_axes#Generalization_to_several_dimensions
EDIT:
AFter answer from @caverac
If not changing the axis
$\vec{new} = R_{active} * \vec{old}$
$R_{z}(active)={\begin{bmatrix}\cos \theta &-\sin \theta &0\\[3pt]\sin \theta &\cos \theta &0\\[3pt]0&0&1\\\end{bmatrix}}$
If changing the axis
$\vec{new} = R_{passive} * \vec{old}$
and
$R_{z}(passive)={\begin{bmatrix}\cos \theta &\sin \theta &0\\[3pt]-\sin \theta &\cos \theta &0\\[3pt]0&0&1\\\end{bmatrix}}$
The angle is just negated ( clockwise or counter-clockwise ). Is my understanding correct?
| If you're referring to the sign difference, this seems to be the key sentence in the SE answer:
In 2D graphic libraries the x-axis goes to the right and the y-axis goes down
The wikipedia entry is based on the y-axis going up. The formula is slightly different depending on whether you consider the angle to be clockwise (as in the SE answer) or counter-clockwise (as in the wikipedia page).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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